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4Sum.java
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/*
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<List<Integer>> fourSum(int[] num, int target) {
int N = num.length;
List<List<Integer>> res = new ArrayList<List<Integer>>();
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<Integer> tmp = new ArrayList<Integer>();
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;
}
}