/**

* Time : O(N^2); Space: O(N)

* @tag : Array; Two Pointers;

* @by : Steven Cooks

* @date: May 2, 2015

***************************************************************************

* 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:

* 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)

*

***************************************************************************

* {@link https://leetcode.com/problems/3sum/ }

*/

package _015_3Sum;

import java.util.ArrayList;

import java.util.List;

import java.util.Arrays;

/** see test {@link _015_3Sum.SolutionTest } */

public class Solution {

public List<List<Integer>> threeSum(int[] nums) {

List<List<Integer>> result = new ArrayList<>();

Arrays.sort(nums);

int n = nums.length;

for (int i = 0; i < n; i++) {

// skip duplicates

if (i != 0 && nums[i] == nums[i - 1]) {

continue;

}

// number will serve as the 1st number in triplet candidate

int num = nums[i];

int left = i + 1;

int right = n - 1;

int target = 0 - num;

// try to find two numbers that sum up to target

while (left < right) {

int twoSum = nums[left] + nums[right];

if (twoSum < target) {

left++;

} else if (twoSum > target) {

right--;

} else {

// found one combination

result.add(Arrays.asList(num, nums[left], nums[right]));

while (left + 1 < right && nums[left] == nums[left + 1]) {

left++;

}

left++;

while (right - 1 > left && nums[right] == nums[right - 1]) {

right--;

}

right--;

}

}

}

return result;

}

}