4Sum.h

/*
 Author:     Annie Kim, anniekim.pku@gmail.com
 Date:       Apr 20, 2013
 Update:     Sep 26, 2013
 Problem:    4Sum
 Difficulty: Medium
 Source:     http://leetcode.com/onlinejudge#question_18
 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.
 */

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        int N = num.size();
        vector<vector<int> > res;
        if (N < 4) return res;
        sort(num.begin(), num.end());
        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) {
                        vector<int> quadruplet(4);
                        quadruplet[0] = num[i];
                        quadruplet[1] = num[j];
                        quadruplet[2] = num[l];
                        quadruplet[3] = num[r];
                        res.push_back(quadruplet);
                        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;
    }
};