Leetcode#1-Two Sum

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 the same element twice.

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

solution：

class Solution {    public int[] twoSum(int[] nums, int target) {        Map<Integer, Integer> map = new HashMap<>();//声明hash map，名字叫做map        for(int i = 0; i < nums.length; i++){        //用for循环把nums里的键值对加入到map中            map.put(nums[i], i); //用.put函数执行加入操作        }        for(int i = 0; i < nums.length; i++){//用for循环检查键值对是否在map中存在            int complement = target - nums[i];            if(map.containsKey(complement) && map.get(complement) != i){//Beware that the complement must not be nums[i]itself!                //比如[2，2，5]，target = 4，此时2的index可以是0也可以是1，i就会有问题。所以为了避免重复，complement的index必须不是nums[i]的index.                //用.comtainsKey函数检查是否complement在map中存在.用.get获取complement的索引值。                return new int[] {i, map.get(complement)};//return新的指定数组，用new int[]{a, b};            }    }    throw new IllegalArgumentException("No two sum solution");//异常处理，throw new IllegalArgumentException}}