/* Author: King, [email protected] Date: Dec 20, 2014 Problem: 3Sum Difficulty: Medium Source: https://oj.leetcode.com/problems/3sum/ Notes: 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) Solution: Simplify '3sum' to '2sum' O(n^2). http://en.wikipedia.org/wiki/3SUM */ public class Solution { public List> threeSum(int[] num) { List> res = new ArrayList>(); Arrays.sort(num); int N = num.length; for (int i = 0; i < N-2 && num[i] <= 0; ++i) { if (i > 0 && num[i] == num[i-1]) continue; // avoid duplicates int twosum = 0 - num[i]; int l = i + 1, r = N - 1; while (l < r) { int sum = num[l] + num[r]; if (sum < twosum) ++l; else if (sum > twosum) --r; else { ArrayList tmp = new ArrayList(); tmp.add(num[i]); tmp.add(num[l]); tmp.add(num[r]); res.add(tmp); ++l; --r; while (l < r && num[l] == num[l-1]) ++l; // avoid duplicates while (l < r && num[r] == num[r+1]) --r; // avoid duplicates } } } return res; } }