# leetcode-sum3

Given an array S of n integers, are there elements a, b, c in S such that
a + b + c = 0?
Find all unique triplets in the array which gives the sum of zero.

Note:

Elements in a triplet (a,b,c) must be in non-descending order.
(ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.

For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:
(-1, 0, 1)
(-1, -1, 2)

```package leetcode

import (
"fmt"
"sort"
"sync/atomic"
)

var arrLen int
var ops uint64 = 0
var sortArr []int

func getArr() []int {
inputArr := []int{-7, -8, -1, -2, -4, 1, 4, 9, 3, 6}
return inputArr
}

func ThreeSum() {
// call get array func
inputArr := getArr()
//make sure are sorted
fmt.Println(inputArr)
if !sort.IntsAreSorted(inputArr) {
// sort inputArr
sort.Ints(inputArr)
fmt.Println(inputArr)
sortArr = inputArr
}
//get this array length
arrLen = len(sortArr)
for i := 0; i < arrLen-2; i++ {
fmt.Println("loop time : ", i)
if i > 0 && sortArr[i-1] == sortArr[i] {
//if near equal ,skip this time loop
continue
}
currNum := sortArr[i]
low := i + 1
high := arrLen - 1
for low < high {
lowNum := sortArr[low]
highNum := sortArr[high]
sumNum := currNum + lowNum + highNum
if sumNum == 0 {
//2d array use
//print result
fmt.Println("--------------Begin--------------")
fmt.Println("counter : ", opsFinal)
fmt.Println("first Number : ", currNum)
fmt.Println("second Number : ", lowNum)
fmt.Println("third Number : ", highNum)
fmt.Printf("[%d,%d,%d]", currNum, lowNum, highNum)
fmt.Println("")
fmt.Println("---------------End---------------")
low++
} else if sumNum > 0 {
for high > 0 && sortArr[high] == sortArr[high-1] {
high--
}
high--
} else {
for low < arrLen && sortArr[low] == sortArr[low+1] {
low++
}
low++
}
}
}
}

```