LeetCode018-4Sum

4Sum

Question:

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example1:

Given array nums = [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]
]

解法代码

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

public class LeetCode18 {

    public static void main(String[] args) {
        int[] nums1 = new int[]{1, 0, -1, 0, -2, 2};
        System.out.println(new LeetCode18().fourSum(nums1, 0).toString());
        
        int[] nums2 = new int[]{};
        System.out.println(new LeetCode18().fourSum(nums2, 0).toString());
        
    }
    
    public List<List<Integer>> fourSum(int[] nums, int target) {
        // 如果请求数组是null或者长度不足4位，则返回一个空List
        if(nums == null || nums.length < 4) {
            return new ArrayList<List<Integer>>();
        }
        List<List<Integer>> tarList = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        for(int i = 0; i < nums.length - 3; i++) {
            int threeSumTar = target - nums[i];
            // 去除重复解
            if(i > 0 && nums[i] == nums[i -1]) {
                continue;
            }
            for(int j = i + 1; j < nums.length - 2; j++) {
                // 去除重复解
                if(j > i + 1 && nums[j] == nums[j -1]) {
                    continue;
                }
                int twoSumTar = threeSumTar - nums[j];
                int left = j + 1;
                int right = nums.length - 1;
                while(left < right) {
                    int twoSum = nums[right] + nums[left];
                    if(twoSumTar == twoSum) {
                        // 添加解
                        tarList.add(Arrays.asList(nums[i], 
                                nums[j], nums[left], nums[right]));
                        // 去除重复解
                        while(left < right && nums[right - 1] == nums[right]) {
                            right--;
                        }
                        while(left < right && nums[left] == nums[left + 1]) {
                            left++;
                        }
                        // 找到解后，将左指针右移一位，右指针左移一位，继续求解
                        left++;right--;
                    } else if(twoSum > twoSumTar) {
                        right--;
                    } else {
                        left++;
                    }
                }
            }
        }
        return tarList;
    }
}

Output：

[[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
[]

Time And Space Complexity

Time: 需要三次循环遍历
Space: 不需要使用额外的存储空间

Tips

• 解法思路与三数之和类似LeetCode015-3Sum