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:...
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 ...
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=...
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) ...
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...
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。
函数部分 ` public class Solution { public int[] twoSum(int[] nums, int target) { int [] answer = new int[2]; int answer_one = 0; int answer_two = 0; int end = 0; for(int i = 0 ;i<nums.length; i++) { answer_one = i; ...
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}...
Solution Brute Force public class Solution { public int[] twoSum(int[] numbers, int target) { int[] res = {-1, -1}; if (numbers.length < 2) return res; for (int i = 0; i < numbers.length; i++) { for (int j = i+1; j < numbers.length; j++) { ...