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LeetCode: 4sum

2024-07-19 06:44:38   222人阅读

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:

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


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