CODE 116: 3Sum

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)

public ArrayList<ArrayList<Integer>> threeSum(int[] num) {// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.if (num.length <= 2) {ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();return results;}ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();Arrays.sort(num);for (int i = 0; i < num.length - 2; i++) {int sum = num[i];int start = i + 1;int end = num.length - 1;while (start < end) {if (num[start] + num[end] > 0 - sum) {while (start < end && num[end - 1] == num[end]) {end--;}end--;} else if (num[start] + num[end] < 0 - sum) {while (start < end && num[start + 1] == num[start]) {start++;}start++;} else {ArrayList<Integer> result = new ArrayList<Integer>();result.add(num[i]);result.add(num[start]);result.add(num[end]);if (!results.contains(result)) {results.add(result);}start++;end--;}}while (i < num.length - 2 && num[i + 1] == num[i]) {i++;}}return results;}