【Shawn-LeetCode】Two Sum

从今日起开始研究LeetCode，争取在暑假完成30道题，题目不是很难，不过受某些人（才不告诉你是谁），打算开始考虑时间复杂度和空间复杂度，大家有兴趣的也可以进行尝试。

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].

我采用了简单粗暴的方法，对它直接进行了筛选：

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

这种复杂度大概是O（n^2）并不是很好，其余复杂度，参考如下：

https://liwyno.github.io/2017/05/28/Leetcode-Algorithms-1/

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