LeetCode 18.四数之和

跟三数之和的解法差不多，只是多了一重for循环，现在时间复杂度为O（n^3）。

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class LeetCode18 {

    public static void main(String[] args) {
        List<List<Integer>> res = new ArrayList<>();
        res = fourSum(new int[]{1,0,-1,0,-2,2}, 0);
        System.out.println("success");
    }

    public static List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums != null && nums.length > 3) {
            Arrays.sort(nums);
                for (int i = 0; i < nums.length - 3; ) {
                    int stepRes = target - nums[i];
                    for (int w = i + 1; w < nums.length - 2; ){
                    int j = w + 1;
                    int k = nums.length - 1;

                    while (j < k) {
                        if (nums[j] + nums[k] == -nums[w] + stepRes) {
                            List<Integer> list = new ArrayList<>(3);
                            list.add(nums[i]);
                            list.add(nums[w]);
                            list.add(nums[j]);
                            list.add(nums[k]);
                            result.add(list);

                            k--;
                            j++;

                            while (j < k && nums[j] == nums[j - 1]) {
                                j++;
                            }
                            while (j < k && nums[k] == nums[k + 1]) {
                                k--;
                            }
                        } else if (nums[j] + nums[k] > -nums[w] + stepRes) {
                            k--;
                            while (j < k && nums[k] == nums[k + 1]) {
                                k--;
                            }
                        } else {
                            j++;
                            while (j < k && nums[j] == nums[j - 1]) {
                                j++;
                            }
                        }
                    }
                    w++;
                    while (w < nums.length - 2 && nums[w] == nums[w - 1]) {
                        w++;
                    }
                }
                i++;
                    while (i < nums.length - 3 && nums[i] == nums[i - 1]) {
                        i++;
                    }
            }
        }
        return result;
    }
}