LeetCode算法入門- 4Sum -day11

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.

Example:

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]
]

給出一個(gè)數(shù)組和Target，讓我們列出不重復(fù)的答案：

class Solution {public List<List<Integer>> fourSum(int[] nums, int target) {int len = nums.length;Arrays.sort(nums);List<List<Integer>> result = new ArrayList<>();//防止重復(fù)可以使用HashSet結(jié)構(gòu)來解決HashSet<List<Integer>> set = new HashSet<>();//定義四個(gè)變量，思路很清晰，不難想到for(int i = 0; i < len; i++){for(int j = i + 1; j < len; j++){int k = j + 1;int l = len -1;while(k < l){if(nums[i] + nums[j] + nums[k] + nums[l] == target){List<Integer> temp = Arrays.asList(nums[i],nums[j],nums[k],nums[l]);//判斷有沒有重復(fù)的答案if(!set.contains(temp)){set.add(temp);result.add(temp);}k++;l--;}else if(nums[i] + nums[j] + nums[k] + nums[l] < target){k++;}else if(nums[i] + nums[j] + nums[k] + nums[l] > target){l--;}}}}return result;} }

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