leetcode解题方案--018--4 sum

最新推荐文章于 2021-02-06 10:59:09 发布

原创最新推荐文章于 2021-02-06 10:59:09 发布 · 221 阅读

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本文解析了给定整数数组寻找四个元素之和等于目标值的所有唯一组合的问题。通过排序数组并使用双指针技巧避免重复解，实现了一个高效的解决方案。

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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: 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]
]

分析

没什么可分析的直接看本博客的3sum即可

 public static List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        List<List<Integer>> list = new LinkedList<List<Integer>>();
        for (int i = 0; i < nums.length - 3; i++) {
            if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
                for (int j = i + 1; j < nums.length - 2; j++) {
                    if (j==i+1 || (j>i+1 && nums[j] != nums[j - 1])) {
                        int sum = target - nums[i] - nums[j];
                        int lo = j + 1;
                        int hi = nums.length - 1;
                        while (lo < hi) {
                            if (nums[lo] + nums[hi] == sum) {
                                list.add(Arrays.asList(nums[i], nums[j], nums[lo], nums[hi]));
                                while (lo < hi && nums[lo] == nums[lo + 1]) {
                                    lo++;
                                }
                                while (lo < hi && nums[hi] == nums[hi - 1]) {
                                    hi--;
                                }
                                ;
                                lo++;
                                hi--;
                            } else if (nums[lo] + nums[hi] > sum) {
                                hi--;
                            } else if (nums[lo] + nums[hi] < sum) {
                                lo++;
                            }
                        }

                    }
                }
            }
        }
        return list;

    }