[Leetcode]1. Two Sum

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https://leetcode.com/problems/two-sum/#/description

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 thesame element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].题目翻译：输入一个整形数组，并指定一个整形target，从数组中找出相加等于target的2个元素并返回它们的index。假设每个输入有且只有一个解，且不能重复使用同一个元素。条件：1. 每次输入假设只有1个确定的答案2. 并没有说明数组是有序的强制解决方案：1. 从数组中第一个数开始（N），相继与后面每一位元素相加，直到遍历到最后一个元素，作为一轮2. 每轮从N+1开始，直到N<数组长度-2（最后结尾的2个元素）。public class Solution {      public int[] twoSum(int[] nums, int target) {         if(nums.length < 2) return null;          for(int first=0;first<nums.length-1;first++){                          int second = first+1;             for(;second<nums.length;second++){                 if(nums[first] + nums[second] == target){                     return new int[]{first, second};                 }             }                      }         return null;     } }


随后查看了示例解法，确实比较强势：
Approach #3 (One-pass Hash Table) [Accepted]public int[] twoSum(int[] nums, int target) {    Map<Integer, Integer> map = new HashMap<>();    for (int i = 0; i < nums.length; i++) {        int complement = target - nums[i];        if (map.containsKey(complement)) {            return new int[] { map.get(complement), i };        }        map.put(nums[i], i);    }    throw new IllegalArgumentException("No two sum solution");}

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