/* Author: King, wangjingui@outlook.com Date: Dec 20, 2014 Problem: 4Sum Difficulty: Medium Source: https://oj.leetcode.com/problems/4sum/ 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: Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a <= b <= c <= d) The solution set must not contain duplicate quadruplets. For example, given array S = {1 0 -1 0 -2 2}, and target = 0. A solution set is: (-1, 0, 0, 1) (-2, -1, 1, 2) (-2, 0, 0, 2) Solution: Similar to 3Sum, 2Sum. */ public class Solution { public List> fourSum(int[] num, int target) { int N = num.length; List> res = new ArrayList>(); if (N < 4) return res; Arrays.sort(num); for (int i = 0; i < N; ++i) { if (i > 0 && num[i] == num[i-1]) continue; // avoid duplicates for (int j = i+1; j < N; ++j) { if (j > i+1 && num[j] == num[j-1]) continue; // avoid duplicates int twosum = target - num[i] - num[j]; int l = j + 1, r = N - 1; while (l < r) { int sum = num[l] + num[r]; if (sum == twosum) { ArrayList tmp = new ArrayList(); tmp.add(num[i]); tmp.add(num[j]); tmp.add(num[l]); tmp.add(num[r]); res.add(tmp); while (l < r && num[l+1] == num[l]) l++; // avoid duplicates while (l < r && num[r-1] == num[r]) r--; // avoid duplicates l++; r--; } else if (sum < twosum) l++; else r--; } } } return res; } }