4Sum Solutions in C++

// Source : https://oj.leetcode.com/problems/4sum/
// Author : Hao Chen
// Date   : 2014-07-03



#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector<vector<int> > threeSum(vector<int> num, int target); 

/*
 * 1) Sort the array,
 * 2) traverse the array, and solve the problem by using "3Sum" soultion.
 */

vector<vector<int> > fourSum(vector<int> &num, int target) {
    vector< vector<int> > result;
    if (num.size() < 4) return result;
    sort( num.begin(), num.end() );

    for(int i = 0; i < num.size() - 3; i++) {
        //skip the duplication
        if (i > 0 && num[i - 1] == num[i]) continue;
        vector<int> n(num.begin()+i+1, num.end());
        vector<vector<int> > ret = threeSum(n, target-num[i]);
        for(int j = 0; j < ret.size(); j++) {
            ret[j].insert(ret[j].begin(), num[i]);
            result.push_back(ret[j]);
        }
    }

    return result; 
}

vector<vector<int> > threeSum(vector<int> num, int target) {

    vector< vector<int> > result;
    //sort the array (if the qrray is sorted already, it won't waste any time)
    sort(num.begin(), num.end());

    int n = num.size();

    for (int i = 0; i < n - 2; i++) {
        //skip the duplication
        if (i > 0 && num[i - 1] == num[i]) continue;
        int a = num[i];
        int low = i + 1;
        int high = n - 1;
        while (low < high) {
            int b = num[low];
            int c = num[high];
            if (a + b + c == target) {
                //got the soultion
                vector<int> v;
                v.push_back(a);
                v.push_back(b);
                v.push_back(c);
                result.push_back(v);
                // Continue search for all triplet combinations summing to zero.
                //skip the duplication
                while(low < n && num[low] == num[low + 1]) low++;
                while(high > 0 && num[high] == num[high - 1]) high--;
                low++;
                high--;
            } else if (a + b + c > target) {
                //skip the duplication
                while(high > 0 && num[high] == num[high - 1]) high--;
                high--;
            } else {
                //skip the duplication
                while(low < n && num[low] == num[low + 1]) low++;
                low++;
            }
        }
    }
    return result;
}


int printMatrix(vector< vector<int> > &vv)
{
    for(int i = 0; i < vv.size(); i++) {
        cout << "[";
        for(int j = 0; j < vv[i].size(); j++) {
            cout << " " << vv[i][j];
        }
        cout << "]" << endl;;
    }
}


int main()
{
    int a[] = { 1, 0, -1, 0, -2, 2 };
    vector<int> n(a, a+6);
    int t = 0;
    vector< vector<int> > v = fourSum(n, t);
    printMatrix(v);

    n.clear();
    int b[] = { -1, -5, -5, -3, 2, 5, 0, 4 };
    n.insert(n.begin(), b, b+8);
    t = -7;
    v = fourSum(n, t);
    printMatrix(v);

    return 0;
}

#include <vector>
using std::vector;
#include <algorithm>
#include <unordered_map>
#include <set>

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        if (num.size() < 4) return vector<vector<int>>{};
        std::set<vector<int>> ret;
        std::sort(num.begin(), num.end());
        std::unordered_map<int, vector<std::pair<int, int>>> cache;
        
        for (size_t i=0; i<num.size(); ++i)
            for (size_t j=i+1; j<num.size(); ++j)
                cache[num[i]+num[j]].emplace_back(i, j);
        
        for (const auto &rp : cache) {
            auto found = cache.find(target - rp.first);
            if (found != cache.end())
                for (const auto &low : rp.second)
                    for (const auto &high : found->second)
                        if (low.second < high.first) ret.insert(vector<int>{num[low.first], num[low.second], num[high.first], num[high.second]});
        }
        
        return vector<vector<int>>(ret.cbegin(), ret.cend());
    }
};

/// Source : https://leetcode.com/problems/4sum/
/// Author : liuyubobobo
/// Time   : 2016-12-06

#include <iostream>
#include <vector>
#include <cassert>
#include <stdexcept>

using namespace std;


/// Two pointers
/// Sort the array first.
/// For every different number a and b, try to find a pair (c, d), which a + b + c + d == 0
/// Using this way, we don't need to see whether the triplet is a repeated one
///
/// Time Complexity: O(nlogn) + O(n^3)
/// Space Complexity: O(1)
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {

        int n = nums.size();

        vector<vector<int>> res;
        if(n < 4)
            return res;

        sort(nums.begin(), nums.end());
        for(int i = 0 ; i <= n - 4 ; i = nextNumberIndex(nums, i))
            for (int j = i + 1; j <= n - 3; j = nextNumberIndex(nums, j)) {

                int t = target - nums[i] - nums[j];

                if(nums[j+1] + nums[j+2] > t || nums[n-1] + nums[n-2] < t)
                    continue;

                int p1 = j + 1;
                int p2 = nums.size() - 1;
                if (p1 >= p2)
                    break;

                while (p1 < p2) {
                    if (nums[p1] + nums[p2] == t) {
                        res.push_back({nums[i], nums[j], nums[p1], nums[p2]});

                        p1 = nextNumberIndex(nums, p1);
                        p2 = preNumberIndex(nums, p2);
                    }
                    else if (nums[p1] + nums[p2] < t)
                        p1 = nextNumberIndex(nums, p1);
                    else // nums[p1] + nums[p2] > t
                        p2 = preNumberIndex(nums, p2);
                }
            }

        return res;
    }

private:
    int nextNumberIndex( const vector<int> &nums , int index ){
        for( int i = index + 1 ; i < nums.size() ; i ++ )
            if( nums[i] != nums[index] )
                return i;
        return nums.size();
    }

    int preNumberIndex( const vector<int> &nums , int index ){
        for( int i = index-1 ; i >= 0 ; i -- )
            if( nums[i] != nums[index] )
                return i;
        return -1;
    }
};


void print_vec(const vector<vector<int>>& vec){
    for(const vector<int>& v: vec){
        for(int e: v)
            cout << e << " ";
        cout << endl;
    }
}

int main() {

    vector<int> nums1 = {1, 0, -1, 0, -2, 2};
    int target1 = 0;
    print_vec(Solution().fourSum(nums1, target1));

    return 0;
}

/// Source : https://leetcode.com/problems/4sum/
/// Author : liuyubobobo
/// Time   : 2017-09-27
#include <iostream>
#include <vector>
#include <cassert>
#include <stdexcept>
#include <unordered_map>
#include <algorithm>

using namespace std;


/// Using hash table
/// Sort the array first.
/// Store every different c + d == t first
/// For every different number a and b, try to find a pair (c, d), which a + b + c + d == 0
///
/// Time Complexity: O(nlogn) + O(n^2) + O(n^3)
/// Space Complexity: O(n^2)
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {

        int n = nums.size();

        vector<vector<int>> res;
        if(n < 4)
            return res;

        sort(nums.begin() , nums.end());

        unordered_map<int, vector<pair<int, int>>> hashtable;
        for(int i = 0 ; i < n - 1 ; i = nextNumberIndex(nums, i))
            for(int j = i + 1 ; j < n ; j = nextNumberIndex(nums, j))
                hashtable[nums[i]+nums[j]].push_back(make_pair(nums[i], nums[j]));

        for( int i = 0 ; i <= n - 4 ; i = nextNumberIndex(nums, i) )
            for (int j = i + 1; j <= n - 3; j = nextNumberIndex(nums, j)) {

                int t = target - nums[i] - nums[j];
                if( nums[j+1] + nums[j+2] > t || nums[n-1] + nums[n-2] < t)
                    continue;

                if(hashtable.find(t) == hashtable.end())
                    continue;

                vector<pair<int,int>>::iterator iter =
                        lower_bound(hashtable[t].begin(), hashtable[t].end(), make_pair(nums[j+1], nums[j+1]));
                for(; iter != hashtable[t].end() ; iter ++)
                    res.push_back({nums[i], nums[j], iter->first, iter->second});
            }

        return res;
    }

private:
    int nextNumberIndex( const vector<int> &nums , int index ){
        for( int i = index + 1 ; i < nums.size() ; i ++ )
            if( nums[i] != nums[index] )
                return i;
        return nums.size();
    }

    int preNumberIndex( const vector<int> &nums , int index ){
        for( int i = index-1 ; i >= 0 ; i -- )
            if( nums[i] != nums[index] )
                return i;
        return -1;
    }
};


void print_vec(const vector<vector<int>>& vec){
    for(const vector<int>& v: vec){
        for(int e: v)
            cout << e << " ";
        cout << endl;
    }
}

int main() {

    vector<int> nums1 = {1, 0, -1, 0, -2, 2};
    int target1 = 0;
    print_vec(Solution().fourSum(nums1, target1));

    return 0;
}