public class Two_Sum_II { public int[] twoSum(int[] numbers, int target) { int[] result = new int[2]; //这里用二分搜索,我常用start和end来命名两头,middle是中间。 int start = 0; int end = numbers.length-1; //这个while循环条件很巧妙,二
classSolution {publicint[] twoSum(int[] nums,inttarget) { HashMap<Integer,Integer> map =newHashMap<Integer, Integer>();int[] ans =newint[2];//合并两个for循环for(inti = 0; i <= nums.length - 1; i++){intt = target -nums[i];if(map.containsKey(t) && map.get(t) !=i){ a...
针对2Sum,先对数组排序,然后使用双指针匹配可行解就可以解决,虽然可以考虑使用HashMap加速搜索,但是对于本题使用HashMap的与否的时间复杂度都一样,都是O(nlog(n))。可以参考这个链接: 求和问题总结(leetcode 2Sum, 3Sum, 4Sum, K Sum),写的很清楚。 这里我只写了2Sum和3Sum的代码,注意要避免重复排序,同时避...
LeetCode 001 Two Sum - Java 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. Example: Given nums = [2, 7, 11, 15]...
[LeetCode] 1. 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) Leet...
LeetCode刷题Two Sum 🍀题目 给定一个整数数组 nums 和一个整数目标值 target,请你在该数组中找出 和为目标值 target 的那 两个 整数,并返回它们的数组下标。 你可以假设每种输入只会对应一个答案。但是,数组中同一个元素在答案里不能重复出现。
LeetCode_1. Two Sum_Solution,原题链接原题中文链接一、题目描述二、题目分析1,常规解法这道题目的意思是给定一个数组和一个值,要求出这个数组中两个值的和等于这个给定值target。输出是有要求的:坐标较小的放在前面,较大的放在后面。这俩坐标不能为零。因此我们可以
Can you solve this real interview question? Two Sum - Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may n
代码(java): 代码语言:txt AI代码解释 class Solution { public int[] twoSum(int[] numbers, int target) { int[] res = new int[2]; int i = 0, j = numbers.length - 1,temp;//i为左指针,j为右指针 while (i < j) { temp=numbers[i] + numbers[j];//先记录两数之和 ...
代码(Java) 代码语言:javascript 代码运行次数:0 运行 AI代码解释 publicclassSolution{publicintgetSum(int a,int b){if(b==0)returna;int sum,up;sum=a^b;up=(a&b)<<1;returngetSum(sum,up);}} 代码很简单,递归调用,在每次调用中都先检查进位的计算是不是为0了,也就是是不是没有进位了,如果没...