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.

class Solution {
public int[] twoSum(int[] nums, int target) {
    int[] result = new int[2];
     A:for(int i=0;i<nums.length-1;i++){
         for(int j=i+1;j<nums.length;j++){
             if(target == nums[i]+nums[j]){
                 result[0] = i;
                 result[1] = j;
                 break A;
             }
         }
     }
    return result;
}
}

在网上看到，这种基本的方法虽然能解决问题，但是不够快捷，也不够逼格，看见有人使用hashMap来解决这个问题，小白没有学过hashmap，所以，借此机会，顺便学习一下。使用hashmap将原本算法的时间复杂度由O(N)降为O(1),key存放数值，value存放出现的位置，从前向后进行遍历，用目标值减去当前的值，看是否存在map中，若存在则取出相应的标号，退出。
Best Practice:

public int[] twoSum(int[] nums, int target) {
int[] answer = new int[2];
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; ++i){
    map.put(nums[i], i);
}
for (int i = 0; i < nums.length; ++i){
    int b = target - nums[i];
      if (map.containsKey(b) && i != map.get(b))
          return new int[]{i, map.get(b)};
 }
 return answer;
}