push_back(dict[query]); break; } } return result; } }; // 下面是测试 int main() { Solution sol; vector<int> arr1={3,2,2,2,2,2,4}; vector<int> arr2={3,2,4}; vector<int> res1= sol.twoSum(arr1, 6); vector<int> res2= sol.twoSum(arr2, 6); for(int i:res1) ...
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> res(2); // 双指针, 先固定一个 for (int i = 0; i < nums.size(); i++) { for (int j = i + 1; j < nums.size(); j++) { if (nums[i] + nums[j] == target) { res[0] =...
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...
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){ if(sortednums[i] + sortednums[j] > target) j--; else...
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: # 创建一个字典,用于记录每个元素最后出现的位置 num_dict = {num: idx for idx, num in enumerate(nums)} # 遍历列表nums中的每个数及其索引 for idx, num in enumerate(nums): # 计算当前数与目标值之间的差值 di...
四、C代码 AI检测代码解析 int* twoSum(int* nums, int numsSize, int target) { int *a = (int*)malloc(2 * sizeof(int)); for(int i = 0;i < numsSize;i++){ for(int j = i + 1;j < numsSize;j++){ if(nums[j] == target - nums[i]){ ...
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()) ...
这次我分别使用 C 语言和 C++ 语言来进行完成。 C ++ 给出的类定义如下: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution { 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...
The function twoSum should return indices of the two numbers suchthat they add up to the target, where index1 must be less than index2.Please note that your returned answers (both index1 and index2) are notzero-based. You may assume that each input would have exactly one solution. Input...
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++) { ...