https://leetcode-cn.com/problems/two-sum/solution/liang-shu-zhi-he-by-leetcode-2/ 一、暴力法(java实现) 暴力法很简单,遍历每个元素x,并查找是否存在一个值与target−x 相等的目标元素。 classSolution {publicint[] twoSum(int[] nums,inttarget) {for(inti = 0; i < nums.length; i++) {f...
Two Sum Java解决方案 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 element twice. Given nums = [2, 7, 11, 15], target = 9, ...
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=...
题目来源：力扣（LeetCode）https://leetcode-cn.com/problems/two-sum 二、解答(java)：方案一：循环遍历，逐个尝试,时间复杂度为O(n^2)class Solution { public int[] twoSum(int[] nums, int target) { int a=0; int b=0; for(int i=0;i<nums.length;i++){ for(int j=i+...
value < b.value; } class Solution { public: vector<int> twoSum(vector<int> &nums, int target) { int len = nums.size(); assert(len >= 2); vector<int> ret(2, 0); // 初始化:ret包含2个值为0的元素 vector<Node> nums2(len); for(int i = 0; i < len; i++){ nums2[i]...
顺道附上java版本的解法: publicclassSolution{publicint[]twoSum(int[]nums,inttarget){int[]res=newint[2];Map<Integer,Integer>map=newHashMap<>();for(inti=0;i<nums.length;i++){if(map.containsKey(nums[i])){res[0]=map.get(nums[i]);res[1]=i;break;}map.put(target-nums[i],i);}re...
C.LeetCode上用时最少的Java Solution 直接来分析这个用时3ms的solution吧 /** * * <p> * Description:截至此刻leetCode上的最优解<br /> * 耗时:3ms<br /> * </p> * 已发现2个无法通过的TestCase:<br /> * 1.Input:[2222222,2222222],4444444;Output:[0,1]<br /> ...
Java解决方案 public class ListNode { int val; ListNode next; ListNode() {} ListNode(int val) { this.val = val; } ListNode(int val, ListNode next) { this.val = val; this.next = next; } } class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummyHead...
class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode head = null; ListNode point = null; int temp = 0; while(l1 != null || l2 != null){ //短的链表空了,就取0,不影响计算 int n1 = l1 == null ? 0:l1.val; ...
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>(); ...