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:

Because nums[0] + nums[1] = 2 + 7 = 9,

• Learn how to use sort in golang

• Solution1: search

In a sorted array, find a[i] + a[j] == target

when a[beg] + a[end] > target, we have two options:

1. beg-- we have already tried a[beg-1] + a[end], which not equal to target, can ignore

2. end-- go ahead

when a[beg] + a[end] < target

1. beg++ go ahead

2. end++ tried before

• Solution2: Using Map

import "sort"type Node struct {    Value int    Index int}type ByVal []*Nodefunc (byVal ByVal) Len() int {return len(byVal)}func (byVal ByVal) Swap(i, j int)  { byVal[i], byVal[j] = byVal[j], byVal[i] }func (byVal ByVal) Less(i, j int) bool { return byVal[i].Value < byVal[j].Value }func twoSum(nums []int, target int) []int {        nodeList := make([]*Node, 0)    for index, num := range nums {        nodeList = append(nodeList, &Node{Value:num, Index:index})    }        sort.Sort(ByVal(nodeList))    //fmt.Printf("%v\n", nodeList)        res := make([]int, 0)    beg := 0;     end := len(nodeList) - 1;    for beg < end {        cur := nodeList[beg].Value + nodeList[end].Value        if cur == target {            res = append(res, nodeList[beg].Index)            res = append(res, nodeList[end].Index)            break        }        if cur > target {            end--        }         if cur < target {            beg++        }     }    sort.Ints(res)    return res}