1. Two Sum Java Solutions Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would haveexactlyone solution. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 ...
publicint[] twoSum(int[] nums,inttarget) {for(inti=0; i < nums.length; i++) {for(intj=i +1; j < nums.length; j++) {if(nums[j] == target - nums[i]) {returnnewint[] { i, j }; } } }thrownewIllegalArgumentException("No two sum solution"); } 复杂度分析: 时间复杂度:...
import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class Solution { //javascript:void(0) //K sum 可以递归去做 /* * 2Sum问题的求解:排序外加双指针来实现 * */ public List<List<Integer>> twoSum(int[] nums,int target) { List<List<Integer>> twoResList=...
public class Solution { public int[] twoSum(int[] nums, int target) { //创建一下数组,要存两个index的数组。 int[] result = new int[2]; //这里用hashtable也行,看心情。 Map<Integer, Integer> map = new HashMap<Integer, Integer>(); //扫一遍数组,一边扫一边存 for(int i = 0; i ...
Two Sum 两数之和 1.Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use ...【Leetcode】1. 两数之和(Two Sum) Leetcode - 1 Two ...
更多文章查看个人博客 个人博客地址:twoSum 两数之和 【JAVA实现】 方法一 使用双重循环两两相加判断是否等于目标值 public List<String> twoSum2(int[] arr, int sum) { if (arr == null || arr.length == 0) { return new ArrayList<>(); } List<String> list = new ArrayList<>(); for (int...
Solve two sum problem (Javascript, Java, C#, Swift, Kotlin, Python, C++, Golang) Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same el...
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) ...
1// 对撞指针2// 时间复杂度: O(n)3// 空间复杂度: O(1)4class Solution{5public:6vector<int>twoSum(vector<int>&numbers,int target){7int l=0,r=numbers.size()-1;8while(l<r){9if(numbers[l]+numbers[r]==target){10int res[2]={l+1,r+1};11returnvector<int>(res,res+2);12}...
这样sum中就储存了所有加入数字可能组成的和,每次find只要花费 O(1) 的时间在集合中判断一下是否存在就行了,显然非常适合频繁使用find的场景。 三、总结 对于TwoSum 问题,一个难点就是给的数组无序。对于一个无序的数组,我们似乎什么技巧也没有,只能暴力穷举所有可能。