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; } } }returna; } ...
index)-1; break; } else if(sum < target){ l++; } else { r--; } } return ret; // 用两个指针来扫 } }; // 下面是测试代码 /* int main() { Solution sol; vector<int> arr; arr.push_back(3); arr.push_back(2); arr.push_back(4); vector<int> ret = sol.twoSum(arr, ...
} 发现对于leetcode而言貌似只能提交class Solution内部的部分。而compare函数放在Solution内部就会出错,放在Solution外面就正常。为什么??? 然后就通过重载圆括号写了函数对象,顺利通过。 1classSolution {//two sum O(nlogn)2public:3structNode{4intval;5intindex;6Node(intv,intidx) :val(v), index(idx){}7...
## 精简版 class Solution: def twosum2(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): ## 遍历 nums,到i for j in range(i+1, len(nums)): ## 从 i 的右边寻找符合条件的元素 if nums[i] + nums[j] == target: return [i, j] ## 保存两个元...
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] =...
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()) ...
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashtable = dict() for i, num in enumerate(nums): if target - num in hashtable: return [hashtable[target - num], i] hashtable[nums[i]] = i ...
语句return returnArr;不返回数组。当一个数组在C中的表达式中使用时，除了作为sizeof的操作数、&的操作...
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++) { //算差值 int resul...
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...