题目

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

有三种解题思路，但是我只谢了两种
第一种当然是最简单的两次循环暴力破解
第二种是将所有的键存储到一个map中以值为键，在存放前先进行一次计算，假定当前值为x,未知值为y，target-x=y，再从map中尝试获取键为y的值，若能获取到则直接返回x的键及map中y的值

答案

1.

func twoSum(nums []int,target int) []int {
    for i:=0;i<len(nums);i++{
        for j:=i+1;j<len(nums);j++{
            if nums[i]+nums[j] == target {
                return []int{i,j}
            }
        }
    }
    return nil
}
//时间复杂度为o(n^2)

2.

func twoSum(nums []int,target int) []int {
    maps := map[int]int{}
    for index,value := range nums {
        member := target - value
        if key,ok := maps[member];ok{
            return []int{key,index}
        }else{
            maps[value] = index   
        }
    }
    return nil
}
//时间复杂度为o(n)