Leetcode: 4Sum

Given an arraySOfNIntegers, are there elementsA,B,C, AndDInSSuch thatA+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:

public class Solution {public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {Arrays.sort(num); HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); for (int i = 0; i < num.length; i++) {for (int j = i + 1; j < num.length; j++) {int k = j + 1;int l = num.length - 1; while (k < l) {int sum = num[i] + num[j] + num[k] + num[l]; if (sum > target) {l--;} else if (sum < target) {k++;} else if (sum == target) {ArrayList<Integer> temp = new ArrayList<Integer>();temp.add(num[i]);temp.add(num[j]);temp.add(num[k]);temp.add(num[l]); if (!hashSet.contains(temp)) {hashSet.add(temp);result.add(temp);} k++;l--;}}}} return result;}}

C ++

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        // Note: The Solution object is instantiated only once and is reused by each test case.        vector<int> tmp;        vector<vector<int>> res;        if(num.empty()) return res;        sort(num.begin(), num.end());        for(int i=0; i<num.size(); i++)        {            int cur = target - num[i];            for(int j=i+1; j<num.size(); j++)            {                int temp = cur - num[j];                int start = j+1, end = num.size()-1;                while(start<end)                {                    if(num[start]+num[end]==temp)                    {                        tmp.push_back(num[i]);                        tmp.push_back(num[j]);                        tmp.push_back(num[start]);                        tmp.push_back(num[end]);                        res.push_back(tmp);                        tmp.clear();                        start++;                        end--;                        while(start<end&&num[start]==num[start-1]) start++;                        while(start<end&&num[end]==num[end+1]) end--;                    }                    else if(num[start]+num[end]<temp)                    {                        start++;                        while(start<end&&num[start]==num[start-1]) start++;                    }                    else                    {                        end--;                        while(start<end&&num[end]==num[end+1]) end--;                    }                }                while(j<num.size()&&num[j]==num[j+1]) j++;             }            while(i<num.size()&&num[i]==num[i+1]) i++;        }        return res;    }};

　　

Leetcode: 4Sum