18. 4Sum

题目:

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)

链接：http://leetcode.com/problems/4sum/

题解: 

同3Sum。 Time Complexity - O(n³)， Space Complexity - O(1)

public class Solution {

    public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {

        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

        if(num == null || num.length < 4)

            return result;

        Arrays.sort(num);

        

        for(int i = 0; i < num.length - 3; i ++){

            if(i > 0 && num[i] == num[i - 1])

                continue;

                

            for(int j = i + 1; j < num.length - 2; j ++){

                if(j > i + 1 && num[j] == num[j - 1])

                    continue;

                    

                int left = j + 1, right = num.length - 1;

                

                while(left < right){

                    int sum = num[i] + num[j] + num[left] + num[right];

                    if(sum == target){

                        ArrayList<Integer> list = new ArrayList<Integer>();

                        list.add(num[i]);

                        list.add(num[j]);

                        list.add(num[left]);

                        list.add(num[right]);

                        result.add(list);

                        left ++;

                        right --;

                        while(left < right && num[left] == num[left - 1])

                            left ++;

                        while(left < right && num[right] == num[right + 1])

                            right --;

                    } else if(sum < target) {

                        left ++;

                    } else {

                        right--;

                    }

                }                

            }

        }
        

        return result;

    }

}