经典算法面试题系列 （二）——three sum

接着two sum，three sum的难度上升了不少。因为two sum的答案是唯一的，three sum的答案是多个。题目链接https://leetcode.com/problems/3sum/#/description。

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: 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]]

其实还是像two sum，先假设已经找到第一个数，剩下的就是循环判断第二个数加上第三个数等于第一个数。当然还有一个问题，就是需要处理重复。处理重复思路是这样的，先找到所有的组合，再做处理。

1.golang 版

package mainimport ("fmt""sort")func main() {var nums = []int{-1, 0, 1, 2, -1, -4}three := threeSum(nums)fmt.Println(three)}func threeSum(nums []int) [][]int {if len(nums) <= 0 {return [][]int{}}if !sort.IntsAreSorted(nums) {sort.Ints(nums)}var sum intvar res [][]intfor i := 0; i < len(nums); i++ {target := -nums[i]front := i + 1back := len(nums) - 1for front < back {sum = nums[front] + nums[back]if sum < target {front++} else if sum > target {back--} else {var tripliet = []int{nums[i], nums[front], nums[back]}res = append(res, tripliet)for front < back && nums[front] == tripliet[1] {front++}for front < back && nums[back] == tripliet[2] {back--}}}for i+1 < len(nums) && nums[i+1] == nums[i] {i++}}return res}

2.php 版

function threeSum($nums) {        if (count($nums) <= 0) {            return [[]];        }        sort($nums);        $res = [];        for($i = 0; $i < count($nums); $i++) {            $target = -$nums[$i];            $front = $i + 1;            $back = count($nums) - 1;            while($front < $back) {                $sum = $nums[$front] + $nums[$back];                if ($sum < $target) {                    $front ++;                } else if ($sum > $target) {                    $back --;                } else {                    $triplet = [$nums[$i], $nums[$front], $nums[$back]];                    $res[] = $triplet;                    while($front < $back && $nums[$front] == $triplet[1]) {                        $front ++;                    }                    while($front < $back && $nums[$back] == $triplet[2]) {                        $back --;                    }                }            }            while($i + 1 < count($nums) && $nums[$i + 1] == $nums[$i]) {                $i ++;            }        }        return $res;    }    $nums = [-1, 0, 1, 2, -1, -4];    $three = threeSum($nums);    var_dump($three);