LeetCode敲門題：

• 題目：

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].
 

• 思路分析：畢竟是第一道題目，很是溫柔，無甚難度，但是直接法的時間見下圖

時間複雜度O(n^2),非常不酷的講，那麼如何優化呢？

小陌是用map屬性，進行邊存邊找

話不多說，看效果先

時間複雜度降到O(n),如果覺得效果還過得去，想瞄幾眼原始碼 ，接著往下瞧：

public class Solution {
	public static void main(String[] args){
        int[] nums = {2,7,11,15};
        int target = 9;
        int[] result = new Solution().twoSum(nums,target);
        for(int i=0;i<result.length;i++){
            System.out.print(result[i]+" ");
        }
    }
    
    public int[] twoSum(int[] nums,int target){
        int[] result = new int[2];
//巧用map的對映，邊存邊找
        HashMap<Integer,Integer> hm = new HashMap<Integer,Integer>();
        for(int i=0;i<nums.length;i++){
            if(hm.get(target-nums[i])!=null){
                result[0] = hm.get(target-nums[i]);
                result[1] = i;
                break;
            }
            else{
                hm.put(nums[i],i);
            }
        }
        return result;
    }
}