腾讯精选练习50题(LeetCode)

发表于

两数相加(medium)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        c1=[]
        c2=[]
        while l1:
            c1.append(l1.val)
            l1=l1.next
        while l2:
            c2.append(l2.val)
            l2=l2.next
        j1=-1
        j2=-1
        sum1=0
        sum2=0
        for i in range(len(c1)):
            sum1=sum1*10+c1[j1]
            j1=j1-1
        for i in range(len(c2)):
            sum2=sum2*10+c2[j2]
            j2=j2-1
        res=sum1+sum2
        p=head=node=ListNode(None)
        l=len(str(res))
        while l:
            node=ListNode(res%10)
            p.next=node
            p=node
            res/=10
            l-=1
        return head.next

寻找两个正序数组的中位数(hard)

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class Solution(object):
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        for i in nums2:
            nums1.append(i)
        nums1 = sorted(nums1)
        if len(nums1)%2 !=0:
            return nums1[len(nums1)/2]
        else:
            return (nums1[len(nums1)/2]+nums1[len(nums1)/2-1])/2.0

最长回文子串(medium)

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char* longestPalindrome(char* s) {
    int i,j,k,t,n,l;
	int count=0;
	int a=0;
    char* ret = (char*)malloc(1001);
    memset(ret,0,1001);
	//int m[1000];
	//char *m=(char *)malloc(sizeof(char)*strlen(s));
	t=strlen(s); 
	for(i=t;i>0;i--){
		for(j=0;j<t-i+1;j++){
			count=0;
			n=j+i-1;
			for(k=j;k<i/2+j;k++){
				if(s[k]==s[n--]){
					count++;
				}
				else
					break;
			}
			if(count==i/2)
				break;
			else
				continue;				
		}
		if(count==i/2)
			break;
		else
			continue;
}
	for(l=j;l<j+i;l++){
		ret[a++]=s[l];
		}
	return ret;
}

整数反转(easy)

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int reverse(int x) {
          int t,sum=0;
	  while(x!=0){
	  	t=x%10;
	  	x=x/10;
        if(sum>2147483647/10||(sum==2147483647/10 && t>7))
              return 0;
        if(sum<-2147483648/10||(sum==-2147483648/10 && t<-8))
              return 0;
	     sum=sum*10+t;
      }
	  return sum;
}

字符串转换整数(medium)

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class Solution(object):
    def myAtoi(self, s):
        """
        :type s: str
        :rtype: int
        """
        res=""
        flag=1
        temp1=["0","1","2","3","4","5","6","7","8","9"]
        temp2=["-","+"]
        temp=0
        for i in range(len(s)):
            if s[i]!=" ":
                temp=i
                break
        s=s[temp:]
        if len(s)==0 or (s[0] not in temp1 and s[0] not in temp2):
            return 0
        elif s[0] in temp1:
            for i in range(len(s)):
                if s[i] in temp1:
                    res += s[i]
                else:
                    break
            if int(res) - 2147483647 > 0:
                return 2147483647
            else:
                return int(res)
        elif s[0] in temp2:
            for i in range(1,len(s)):
                if s[i] in temp1:
                    flag=0
                    res += s[i]
                else:
                    break
            if flag:
                return 0
            elif s[0]=="-":
                if -int(res) - (-2147483648) >= 0:
                    return -int(res)
                else:
                    return -2147483648
            elif s[0]=="+":
                if int(res) - 2147483647 > 0:
                    return 2147483647
                else:
                    return int(res)
        else:
            return 0

回文数(easy)

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bool isPalindrome(int x) {
    int k,t,sum=0;
    if(x<0)
	    return false;
	else
	{
		k=x;
		while(x!=0){
			t=x%10;
			sum=sum*10+t;
			x=x/10;
		}
		if(sum==k)
			return true;
		else
			return false;
	} 
	return 0;
}

盛最多水的容器(medium)

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class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        i=0
        j=len(height)-1
        res=0
        while i<j:
            h=min(height[i],height[j])
            res=max(res,h*(j-i))
            if height[i]>height[j]:
                j-=1
            else:
                i+=1
        return res

最长公共前缀(easy)

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class Solution(object):
    def longestCommonPrefix(self, strs):
        """
        :type strs: List[str]
        :rtype: str
        """
        if len(strs)==0:
            return ""
        c=[]
        for i in range(len(strs)):
            c.append(len(strs[i]))
        minl = min(c)
        s=''
        for i in range(minl):
            res=[]
            for j in range(len(strs)):
                res.append(strs[j][i])
            if res[1:] == res[:-1]:
                s+=strs[j][i]
            else:
                break
        return s

三数之和(medium)

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class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        #第一种方法，可行但超时
        # if len(nums)==0 or len(nums)==1 or len(nums)==2:
        #     return []
        # sum_two=[]
        # d={}
        # res=[]
        # count=0
        # for i in range(len(nums)-1):
        #     for j in range(i+1,len(nums)):
        #         d[count]=[i,j]
        #         sum_two.append(nums[i]+nums[j])
        #         count+=1
        # # print(d)
        # # print(sum_two)
        # for i in range(len(nums)):
        #     for j in range(len(sum_two)):
        #         if i not in d[j] and nums[i]+sum_two[j]==0:
        #             temp = [nums[d[j][0]]] + [nums[i]] + [nums[d[j][1]]]
        #             temp.sort()
        #             if temp not in res:
        #                 res.append(temp)
        # # res = list(set([tuple(t) for t in res]))
        # # res = list([list(l) for l in res])
        # return res

        d = {}
        # n = len(nums)
        res = []
        for i in range(len(nums)):
            d = {}
            for j in range(i + 1, len(nums)):
                key = nums[j]
                if key in d:
                    value = d[key] + [key]
                    value.sort()
                    if value not in res:
                        res.append(value)
                key = -nums[i] - nums[j]
                if key not in d:
                    d[key] = [nums[i], nums[j]]
        return res

最接近的三数之和(medium)

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class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        res=[]
        nums.sort()
        for i in range(len(nums)):
            low=i+1
            high=len(nums)-1
            while low<high:
                sums=nums[i]+nums[low]+nums[high]
                res.append(sums)
                if sums==target:
                    return target
                elif sums<target:
                    low+=1
                else:
                    high-=1
        res.sort()

        if target<=res[0]:
            return res[0]
        if target>=res[-1]:
            return res[-1]
        if target in res:
            return target
        for i in range(1,len(res)):
            if target>res[i-1] and target<res[i]:
                a=target-res[i-1]
                b=res[i]-target
                if a>b:
                    return res[i]
                else:
                    return res[i-1]

有效的括号(easy)

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class Solution(object):
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        c=[]
        j=0
        if s=='':
            return True
        if s[0]==')' or s[0]==']' or s[0]=='}':
            return False
        for i in range(len(s)):
            if len(c)==0:
                c.append(s[i])
            else:
                if (c[-1]=='(' and s[i]==')') or (c[-1]=='[' and s[i]==']')  or (c[-1]=='{' and s[i]=='}'):
                    c.pop(-1)
                else:
                    c.append(s[i])
        if len(c)==0:
            return True
        else:
            return False

合并两个有序链表(easy)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        res=ListNode(None) #合并之后的新数组
        p=res #保持头节点不动
        if l1==None:  
            res.next=l2
            return res.next
        if l2==None:
            res.next=l1
            return res.next
        while l1 and l2:
            if l1.val < l2.val:
                p.next=l1
                l1=l1.next
            else:
                p.next=l2
                l2=l2.next
            p=p.next
        if l1: #如果l1还没循环结束
            p.next=l1
        else: #如果l2还没循环结束
            p.next=l2
        return res.next

合并K个升序链表(hard)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        #如果lists为空，直接返回空
        if len(lists)==0:
            return None
        #temp用来存两个单链表合并后的结果
        temp=ListNode()
        temp.val=None #将初始temp值置空
        #从第一个升序链表开始比较
        for i in range(len(lists)):
            #保存temp和每一个升序链表比较后的结果
            node=ListNode()
            #如果当前链表为空，直接将node指向temp，并开始比较下一个链表
            if lists[i]==None:
                node.next=temp
                continue
            #如果temp为空，node指向当前链表并将temp指向node
            if temp==None:
                node.next=lists[i]
                temp=node.next
                continue
            #两个都不为空，需要一个保存node节点，所以使用head开始向后比较
            head=node
            while temp !=None and lists[i]!=None:
                if temp.val>lists[i].val:
                    head.next=lists[i]
                    head=head.next
                    lists[i]=lists[i].next
                else:
                    head.next=temp
                    head=head.next
                    temp=temp.next
            if temp!=None:
                head.next=temp
            if lists[i]!=None:
                head.next=lists[i]
            #两个链表比较完毕后将temp重置，并开始下一轮的比较
            temp=node.next
        #最后返回node节点
        return node.next.next

删除有序数组中的重复项(easy)

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class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        c=0
        for i in range(1,len(nums)):
            if nums[i]!=nums[c]:
                nums[c+1]=nums[i]
                c+=1
        return c+1

搜索旋转排序数组(medium)

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class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        if target in nums:
            return nums.index(target)
        return -1

字符串相乘(medium)

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class Solution(object):
    def multiply(self, num1, num2):
        """
        :type num1: str
        :type num2: str
        :rtype: str
        """
        # return str(int(num1)*int(num2))
        res = 0
        for i in range(1,len(num1)+1):
            for j in range(1, len(num2)+1):
                res += int(num1[-i]) * int(num2[-j]) *10**(i+j-2)
        return str(res)

全排列(medium)

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class Solution(object):
    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        def trace(path,choice):
            if len(path)==len(nums):
                res.append(path[:])
            for i in choice:
                if i in path:
                    continue
                path.append(i)
                trace(path,choice)
                path.pop()
        trace([],nums)
        return res

最大子序和(easy)

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class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums)==1:
            return nums[0]
        for i in range(1,len(nums)):
            nums[i] = nums[i]+max(nums[i-1],0)
        return max(nums)

螺旋矩阵(medium)

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class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        res=[]
        while len(matrix)!=0:
            for i in matrix[0]:
                res.append(i)
            matrix.pop(0)
            matrix=list(zip(*matrix))
            matrix.reverse()
        return res

螺旋矩阵II(medium)

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class Solution(object):
    def generateMatrix(self, n):
        """
        :type n: int
        :rtype: List[List[int]]
        """
        if n==1:
            return [[1]]
        res=[[n*n-1],[n*n]]
        i=n*n-2
        while i>=1:
            for j in range(len(res)):
                res[j].insert(0,i)
                i-=1
            res=list(map(list,zip(*res)))
            for j in res:
                j.reverse()
        return res

旋转链表(medium)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if head==None:
            return head
        node=head
        count=0
        while node!=None:
            count+=1
            pre=node
            node=node.next
        s=k%count
        t=count-s
        pre.next=head
        for i in range(t):
            prenode=head
            head=head.next
        prenode.next=None
        return head

不同路径(medium)

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class Solution(object):
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        res=[[1]*n]*m
        # print(res)
        for i in range(1,m):
            for j in range(1,n):
                res[i][j]=res[i][j-1]+res[i-1][j]
        return res[-1][-1]

爬楼梯(easy)

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class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        res=[1,2,3]
        for i in range(n-3):
            res.append(res[-1]+res[-2])
        return res[n-1]

子集(medium)

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class Solution(object):
    def subsets(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res=[[]]
        for i in nums:
            res+=[j+[i] for j in res]
        return res

合并两个有序数组(easy)

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void merge(int* nums1, int m, int* nums2, int n) {
    int i = m - 1,j = n - 1,k = m + n - 1;
    while(i >= 0 && j >= 0)
    {
        if(nums1[i] > nums2[j])
            nums1[k--] = nums1[i--];
        else
            nums1[k--] = nums2[j--];
    }
    while(j >= 0)
        nums1[k--] = nums2[j--];
}

格雷编码(medium)

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class Solution(object):
    def grayCode(self, n):
        """
        :type n: int
        :rtype: List[int]
        """
        #先右移再异或
        res=[]
        for i in range(2**n):
            res.append(i^i>>1)
        return res

二叉树的最大深度(easy)

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        return max(self.maxDepth(root.left),self.maxDepth(root.right))+1

买卖股票的最佳时机(easy)

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class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        # max_num=0
        # if len(prices)==0:
        #     return 0
        # for i in range(1,len(prices)):
        #     if prices[i]-min(prices[:i])>max_num:
        #         max_num = prices[i]-min(prices[:i])
        # return max_num

        max_num=0
        if len(prices)==0:
            return 0
        min_num=prices[0]
        for i in range(1,len(prices)):
            min_num = min(min_num,prices[i-1])
            if prices[i]-min_num>max_num:
                max_num = prices[i]-min_num
        return max_num

买卖股票的最佳时机II(easy)

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class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        sum=0
        res=[]
        temp=[]
        #这段用来分割连续增加的点
        for i in range(1,len(prices)):
            if prices[i]<prices[i-1]:
                temp.append(i)
        if len(prices) not in temp:
            temp.append(len(prices))
        temp.insert(0,0)

        #这段用来将每个连续增加段存入res
        for i in range(1,len(temp)):
            res.append(prices[temp[i-1]:temp[i]])
        #计算每个段中最大值和最小值并累加
        for i in res:
            if len(i)>=2:
                sum+=(max(i)-min(i))
        return sum

二叉树的最大路径和(hard)

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只出现一次的数字(easy)

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class Solution:
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        a = 0
        for num in nums:
            a = a ^ num
        return a

环形链表(easy)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if head==None or head.next==None:
            return False
        p=head
        q=head.next
        while p:
            if q.next==p:
                return True
            else:
                if q.next==None:
                    return False
                if q.next.next==None:
                    return False
                p=p.next
                q=q.next.next
        return False

环形链表II(medium)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        res=[]
        node=head
        while node!=None:
            if node.next not in res:
                res.append(node)
                node=node.next
            else:
                return node.next
        return None

LRU缓存机制(medium)

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class LRUCache(object):

    def __init__(self, capacity):
        """
        :type capacity: int
        """
        self.capacity = capacity
        self.d = collections.OrderedDict()


    def get(self, key):
        """
        :type key: int
        :rtype: int
        """
        if key not in self.d:
            return -1
        res = self.d[key]
        self.d.pop(key)
        self.d[key]= res
        return res


    def put(self, key, value):
        """
        :type key: int
        :type value: int
        :rtype: None
        """
        if key not in self.d:
            if self.capacity <= 0:
                for k in self.d.keys():
                    self.d.pop(k)
                    break
            else:
                self.capacity-=1
        else:
            self.d.pop(key)
        self.d[key]=value
        



# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)

排序链表(medium)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        # h_head = ListNode(None)
        # print(h_head)
        # res = []
        # while head != None:
        #    res.append(head)
        #    head = head.next
        # print(res) 
        # res = sorted(res,key = lambda x:x.val)
        # print(res)
        # n = len(res)
        # h_head.next = res[0]
        # # res[0].next = res[1]
        # return h_head.next
        # for i in range(n-1):
        #     res[i].next = res[i+1]
        # return h_head.next


        h_head = ListNode(None) #申请头节点
        res = [] #将节点保存在列表中
        while head is not None:
            next_h = head.next
            head.next = None
            res.append(head)
            head = next_h
            # head = head.next
        res = sorted(res,key = lambda x:x.val)
        # print(res)
        n = len(res)
        if n == 0:
            return None
        h_head.next = res[0]
        for i in range(n-1):
            res[i].next = res[i+1]
        # print(res)
        return h_head.next

最小栈(easy)

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class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.l = [] #列表操作
        self.index = -1


    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        # self.l.append(x)
        # self.minnum = min(self.l)

        self.l.append(x)
        self.index+=1


    def pop(self):
        """
        :rtype: None
        """
        self.l.pop()
        # self.minnum =not self.l or min(self.l)
        self.index-=1


    def top(self):
        """
        :rtype: int
        """
        return self.l[-1]
        # return self.l[self.index]


    def getMin(self):
        """
        :rtype: int
        """
        # return self.minnum
        return min(self.l)



# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()

相交链表(easy)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        l1=0
        l2=0
        node1=headA
        node2=headB
        while headA: #求第一个链表长度
            l1+=1
            headA=headA.next
        while headB: #求第二个链表长度
            l2+=1
            headB=headB.next
        s=l1-l2
        while s>0: #对齐两个链表长度
            node1=node1.next
            s-=1
        while s<0:
            node2=node2.next
            s+=1
        while node1 and node2:
            if node1==node2:
                return node1
            node1=node1.next
            node2=node2.next
        return None
        # while node1!=node2: #从头开始判断地址是否相同
        #     node1=node1.next
        #     node2=node2.next
        # return node1

多数元素(easy)

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class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = sorted(nums)
        return res[(len(nums)//2)]

反转链表(easy)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        # if head:
        #     pre=head
        #     q=head.next
        #     while pre.next:
        #         pre.next=q.next
        #         q.next=head
        #         head=q
        #         q=pre.next
        # return head
        if head==None or head.next==None:
            return head
        pre=head
        q=head.next
        while pre.next:
            pre.next=q.next
            q.next=head
            head=q
            q=pre.next
        return head

数组中的第K个最大元素(medium)

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class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        s = sorted(nums,reverse=True)
        return s[k-1]

存在重复元素(easy)

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class Solution(object):
    def containsDuplicate(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        d = dict(Counter(nums))
        for v in d.values():
            if v>=2:
                return True
        return False

二叉搜索树中第K小的元素(medium)

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def kthSmallest(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: int
        """
        res = []

        def pre(node):
            if node == None:
                return 0
            pre(node.left)
            res.append(node.val)
            pre(node.right)
        pre(root)
        return res[k-1]
        # print(res)

2的幂(easy)

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bool isPowerOfTwo(int n) {
    if(n<=0)
        return false;
    else
        if((n&n-1)==0)
            return true;
        else
            return false;
}

二叉搜索树的最近公共祖先(easy)

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if root==None or root==p or root==q:
            return root
        left = self.lowestCommonAncestor(root.left,p,q)
        right= self.lowestCommonAncestor(root.right,p,q)
        if left!=None and right!=None:
            return root
        if left!=None:
            return left
        if right!=None:
            return right
        return None

二叉树的最近公共祖先(medium)

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        #如果root不存在或者有一个值是根节点，直接返回根节点即可
        if root == None or root==p or root==q:
            return root
        #递归左右子树查找
        left = self.lowestCommonAncestor(root.left,p,q)
        right = self.lowestCommonAncestor(root.right,p,q)
        if left!=None and right != None:
            return root
        elif left!=None:
            return left
        elif right!=None:
            return right
        return None

删除链表中的节点(easy)

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val=node.next.val
        node.next=node.next.next

除自身以外数组的乘积(medium)

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class Solution(object):
    def productExceptSelf(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        res=[]
        res1=[1 for i in range(len(nums))]
        res2 = [1 for i in range(len(nums))]
        for i in range(1,len(nums)):
            res1[i]=nums[i-1]*res1[i-1]
        for j in range(len(nums)-2,-1,-1):
            res2[j]=nums[j+1]*res2[j+1]
        for i in range(len(res1)):
            res.append(res1[i]*res2[i])
        return res
        # print(res)
        # for i in range(len(nums)):
        #     for j in range(i):
        #         res[i]*=nums[j]
        #     for k in range(i+1,len(nums)):
        #         res[i]*=nums[k]
        # return res

Nim游戏(easy)

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bool canWinNim(int n) {
    if(n%4!=0)
        return true;
    else
        return false;
}

反转字符串(easy)

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void reverseString(char* s, int sSize) {
    int temp;
    int i;
    for(i=0;i<sSize/2;i++){
        temp=s[i];
        s[i]=s[sSize-i-1];
        s[sSize-i-1]=temp;
    }
    return s;
}

反转字符串中的单词III(easy)

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class Solution(object):
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        return " ".join(i[::-1] for i in s.split())

        # res = ""
        # s1 = list(s.split(" "))
        # for i in range(len(s1)-1):
        #     res+=s1[i][::-1]
        #     res+=" "
        # res+=s1[-1][::-1]
        # return res