Two Sum

0x00

难度：Easy
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].

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本题很简单，最暴力的方法是循环遍历数组：

func twoSum(nums []int, target int) []int {
	var ret []int
	for i,vi := range nums {
		for j,vj := range nums[i+1:] {
			if vi + vj == target {
				return append(ret,i,i + j + 1)
			}
		}
	}
	return ret
}

但是我们仔细审题会发现

have exactly one solution, and you may not use the same element twice.

这就意味着如果vi + vj = target，那么只要知道 vj 的值，vi 的值就等于target - vi，而且 vi 与 vj 是唯一的，因此可以将vi,vi作为key，使用map更高效的实现。

func twoSum(nums []int, target int) []int {
	var ret []int
	m := make(map[int]int)
	for i,vi := range nums {
		if j,ok := m[target - vi];ok {
			return append(ret,j,i)
		}
		m[vi] = i
	}
	return ret
}