LeetCode : 4Sum [java]

最新推荐文章于 2021-02-22 20:44:31 发布

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最新推荐文章于 2021-02-22 20:44:31 发布

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分类专栏： Java 编程练习题文章标签： LeetCode 4Sum

本文链接：https://blog.youkuaiyun.com/hackeraaa/article/details/50817563

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本文介绍了一种解决四数之和问题的有效算法。通过先排序数组，再使用双指针技术来寻找所有唯一四元组，使得这些四元组的元素之和等于目标值。该方法避免了重复解，并确保了四元组内的元素按非递减顺序排列。

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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)

思路：确定前两个，双指针扫描后两个。

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

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
		List<List<Integer>> list = new ArrayList<List<Integer>>();
		if (nums == null || nums.length < 4) {
			return list;
		}
		Arrays.sort(nums);
		for (int i = 0; i < nums.length; i++) {
			while (i > 0 && i < nums.length && nums[i] == nums[i - 1]) {
				i++;
			}
			for (int j = i + 1; j < nums.length; j++) {
				while (j > i + 1 && j < nums.length && nums[j] == nums[j - 1]) {
					j++;
				}
				int left = j + 1;
				int right = nums.length - 1;
				while (left < right) {

					int sum = nums[i] + nums[j] + nums[left] + nums[right];
					if (sum > target) {
						while (right > left && nums[right] == nums[right - 1]) {
							right--;
						}
						right--;
					} else if (sum < target) {
						while (left < right && nums[left] == nums[left + 1]) {
							left++;
						}
						left++;
					} else {
						List<Integer> interList = new ArrayList<Integer>();
						interList.add(nums[i]);
						interList.add(nums[j]);
						interList.add(nums[left]);
						interList.add(nums[right]);
						list.add(interList);
						while (left < right && nums[left] == nums[left + 1]) {
							left++;
						}
						left++;
					}
				}
			}
		}
		return list;
    }
}