The problem "Two Sum" requires finding two numbers in aninteger arraysuch that their sum equals a specifiedtargetnumber. You need to returnthe indices ofthese two numbers, whereindices start from 0. The indices ofthe two numbers cannot be the same, and there isexactly one solutionfor each i...
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 ...
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...
5 这一步提供我的打败97%的人的代码实现代码：class Solution {public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int,int > map; for(int i=0;i<nums.size();i++) { int val=nums[i]; auto iter=map.find(val); if (iter!=map.end()) ...
classSolution{public:vector<int>twoSum(vector<int>&nums,inttarget){unordered_map<int,int>num_map;for(inti=0;i<nums.size();i++){autoit=num_map.find(target-nums[i]);// 此处不需要去重判断,按照题目意思,数组中不存在重复数字// 因此在一次遍历的情况下,不会出现重复数字if(it!=num_map.end(...
classSolution { public: vector<int> twoSum(vector<int>& nums,inttarget) { vector<int> result; vector<int> sortednums(nums); sort(sortednums.begin(),sortednums.end()); inti = 0, j = sortednums.size()-1; while(i != j){
classSolution{public:vector<int>twoSum(vector<int>&nums,int target){}}; C++ 类中的 twoSum 成员函数有两个参数,分别是 nums 和 target,这两个参数和题目中描述的是一样的。 C 语言给出的函数定义如下: 代码语言:javascript 复制 /** * Note: The returned array must be malloced, assume caller ca...
classSolution{public:vector<int>twoSum(vector<int>&nums,inttarget){vector<int>vecResult;autoiSize=nums.size();autoiOutterSize=iSize-1;for(inti=0;i<iOutterSize;++i){for(intj=i+1;j<iSize;++j){if(nums[i]+nums[j]==target){vecResult.push_back(i);vecResult.push_back(j);returnvecRe...
代码语言:javascript 复制 classSolution:deftwoSum(self,nums,target):iflen(nums)<=1:returnFalse d=dict()foriinrange(len(nums)):ifnums[i]ind:return[d[nums[i]],i]else:d[target-nums[i]]=i 恩,最后找队友一起刷题。喜欢可以联系我
class Solution { public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap(); //存储到HashMap中 for (int i = 0; i < nums.length; i++){ map.put(nums[i], i); } //遍历数组 for (int i = 0; i < nums.length; i++) { ...