java - 4sum 在 Java 中的实现，来自 leetcode

Question

leetcode 中的问题陈述说:

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.

注意:

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.

关于这个四和问题，我有两个问题:

1. 我哪里错了？编译后的代码无法通过所有测试，但我认为代码应该是正确的，因为它只是使用暴力来解决问题。
2. 是否有任何有效的方法来解决四和问题而不是这个 O(n^4) 运行时算法？

---

import java.util.List;
import java.util.ArrayList;
import java.util.Collections;

public class FourSumSolution{
    public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target){
        ArrayList<ArrayList<Integer>> aList = new ArrayList<ArrayList<Integer>>();
        int N = num.length;

        for(int i=0; i<N; i++)
            for(int j=i+1; j<N; j++)
                for(int k=j+1; k<N; k++)
                    for(int l=k+1; l<N; l++){
                        int sum = num[i] + num[j] + num[k] + num[l];                        
                        if(sum == target){
                            ArrayList<Integer> tempArray = new ArrayList<Integer>();
                            Collections.addAll(tempArray, num[i], num[j], num[k], num[l]);
                            aList.add(tempArray);
                        }
                    }
        return aList;   
    }
}

Answer 1

这是一个 O(n^3) 的解决方案

import java.util.Arrays;
import java.util.ArrayList;
import java.util.HashSet;
public class Solution {
public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
    // Start typing your Java solution below
    // DO NOT write main() function

    Arrays.sort(num);
    HashSet<ArrayList<Integer>> hSet = 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++) {
            for (int k = j + 1, l = num.length - 1; 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> found = new ArrayList<Integer>();
                    found.add(num[i]);
                    found.add(num[j]);
                    found.add(num[k]);
                    found.add(num[l]);
                    if (!hSet.contains(found)) {
                        hSet.add(found);
                        result.add(found);
                    }

                    k++;
                    l--;

                }
            }
        }
    }
    return result;
}