Leetcode 18.4Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/StromMaybin/article/details/78434177

本文探讨了在给定整数数组中寻找四个元素之和等于特定目标值的所有唯一组合的问题。基于三数之和的基础上增加了额外一层循环，并通过剪枝技术减少不必要的计算，最终实现时间复杂度为O(N^3)的解决方案。

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@author stormma
@date 2017/11/03

生命不息，奋斗不止！

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]
]

思路分析

在3sum基础上，再加一次循环即可，可以增加剪枝来换取运行时间，
时间复杂度O(N^3)

代码实现


class Solution {
        public List<List<Integer>> fourSum(int[] nums, int target) {
            List<List<Integer>> ans = new ArrayList<>();
            if (nums.length < 4) {
                return ans;
            }
            Arrays.sort(nums);
            if (4 * nums[0] > target || 4 * nums[nums.length - 1] < target) {
                return ans;
            }
            int max = nums[nums.length - 1];
            // -2 -1 0 0 1 2
            for (int i = 0; i < nums.length - 3; i++) {
                if (i > 0 && nums[i] == nums[i - 1]) {
                    continue;
                }
                if (4 * nums[i] > target) {
                    break;
                }
                // 剪枝
                if (nums[i] + 3 * max < target) {
                    continue;
                }
                for (int j = i + 1; j < nums.length - 2; j++) {
                    // 3
                    // -4 -3 0 5 5 10
                    int low = j + 1, high = nums.length - 1, goal = target - nums[i] - nums[j];
                    // 剪枝
                    if (nums[i] + nums[j] + 2 * max < target) {
                        continue;
                    }
                    // 去重
                    if (j > i + 1 && nums[j - 1] == nums[j]) {
                        continue;
                    }
                    while (low < high) {
                        if (nums[low] + nums[high] < goal) {
                            low++;
                        } else if (nums[low] + nums[high] > goal) {
                            high--;
                        } else {
                            ans.add(Arrays.asList(nums[i], nums[j], nums[low], nums[high]));
                            while (low < high && nums[low + 1] == nums[low]) {
                                low++;
                            }
                            while (low < high && nums[high - 1] == nums[high]) {
                                high--;
                            }
                            low++;
                            high--;
                        }
                    }
                }
            }
            return ans;
        }
    }