result[0] =i result[1] =jreturnresult[:]//返回结果} } }returnnil } 回到顶部 四、C代码 int* twoSum(int* nums,intnumsSize,inttarget) {int*a = (int*)malloc(2*sizeof(int));for(inti =0;i < numsSize;i++){for(intj = i +1;j < numsSize;j++){if(nums[j] == target -nums[i]){ a[0] =i; a[1] =j; } } ...
这个代码完全不能通过LeetCode的测试。算法复杂度太高 然后我看了别人写的 1classSolution(object):2deftwoSum(self,nums,target):3"""4:type nums: List[int]5:type target: int6:rtype: List[int]7"""8n =len(nums)9result ={}10ifn <= 1:11returnFalse12else:13foriinrange(n):14ifnums[i]i...
Can you solve this real interview question? Two Sum - Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may n
这样我们创建一个哈希表,对于每一个 x,我们首先查询哈希表中是否存在 target - x,然后将 x 插入到哈希表中,即可保证不会让 x 和自己匹配。 class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashtable = dict() for i, num in enumerate(nums): if target - num ...
LeetCode_1. Two Sum_Solution,原题链接原题中文链接一、题目描述二、题目分析1,常规解法这道题目的意思是给定一个数组和一个值,要求出这个数组中两个值的和等于这个给定值target。输出是有要求的:坐标较小的放在前面,较大的放在后面。这俩坐标不能为零。因此我们可以
leetcode: Subarray Sums Divisible by K Problem Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Solution 使用prefix sums的方法。 用一个presum变量记录当前位置 i 的总和A[0]+...+A[i]A... hdu 5637 BestCod...
这是每个初次接触leetcode的同学都将做的第一道题。题目本身的思维方式十分简单,可采用暴力破解法,利用for循环嵌套,便可通过测试: class Solution: def twoSum(nums: list, target: int) -> list: newlist = [] for firstIndex in range(0, len(nums)-1): for secondIndex in range(firstIndex+1, len...
leetcode第28题--Divide Two Integers Divide two integers without using multiplication, division and mod operator. 分析:题目意思很容易理解,就是不用乘除法和模运 ... [LeetCode]29. Divide Two Integers两数相除 Given two integers dividend and divisor, divide two integers without using multiplication...
vector<int>twoSum(vector<int>&numbers,int target){vector<int>res;if(numbers.size()<2){returnres;}intleft=0,right=numbers.size()-1;while(left<right){if(target==numbers[left]+numbers[right]){res.push_back(left+1);res.push_back(right+1);returnres;}elseif(target>numbers[left]+numbers...
nums = [2,7,11,15] & target = 9 -> [0,1], 2 + 7 = 9 At each num, calculate complement, if exists in hash map then return Time: O(n) Space: O(n) */ class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { int n = nums.size(); unordered_map...