solution: voidrecursive(int** result,int* nums,intnumsSize,int* returnSize,bool* used,int* temp,intsize){inti=0;if(size==numsSize) { result[*returnSize]=(int*)malloc(sizeof(int)*numsSize);for(i=0;i<numsSize;i++) { result[*returnSize][i]=temp[i]; } size=0; (*returnSize)++;...
https://oj.leetcode.com/problems/permutations/ Given a collection of numbers, return all possible permutations. For example, [1,2,3]have the following permutations: [1,2,3],[1,3,2], [2,1,3],[2,3,1], [3,1,2], and[3,2,1]. 解题思路: 一道枚举出所有组合的题目,和前面的Letter...
与Permutations不一样的地方就是有duplicates。办法就是跳过duplicates(29-33行)。其他不变。 【代码】 public class Solution { public List<List<Integer>> permuteUnique(int[] nums) { //require List<List<Integer>> ans=new ArrayList<>(); if(nums==null) return ans; int size=nums.length; if(size...
LeetCode: 46. Permutations 题目描述 Given a collection of distinct numbers, return all possible permutations. For example, [1,2,3] have the following permutations: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ] 1. 2. 3. 4. 5. 6. ...
leetcode[47] Permutations II Given a collection of numbers that might contain duplicates, return all possible unique permutations. Example: Input: [1,1,2] Output: [ [1,1,2], [1,2,1], [2,1,1] ] 题目大意: 给一个数组含有重复数字,找出他们所有不重复的全排列。
https://leetcode.com/problems/permutations-ii/ Given a collection of numbers that might contain duplicates, return all possible unique permutations. Example: Input: [1,1,2] Output: [ [1,1,2], [1,2,1], [2,1,1] ] 代码: 1
[1,1,2] have the following unique permutations: [ [1,1,2], [1,2,1], [2,1,1] ] 思路 本题思路和上一篇博客Permutations一样,唯一区别就是,加入了去重操作,构建了set容器 代码 AI检测代码解析 class Solution { public: vector<vector<int>> permuteUnique(vector<int>& nums) { ...
https://discuss.leetcode.com/topic/46162/a-general-approach-to-backtracking-questions-in-java-subsets-permutations-combination-sum-palindrome-partioning 另外一个别人的算法,用的是iterative,但是我不准备研究了 https://discuss.leetcode.com/topic/6377/my-ac-simple-iterative-java-python-solution 更多讨论:...
来源: https://leetcode.com/problems/permutations/ 分析 方法一: 因为每个元素只能使用一次,所以使用一个数组,来记录该元素是否已经在上一级被使用过,如果使用过,则跳过,不进行DFS。该方法使用额外的数组来记录nums[i]是否被使用,同时每次的递归调用都会创建新的result,占用额外空间。生成的过程类似与下图:1...
https://leetcode.com/problems/permutations-ii/ Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example, [1,1,2]have the following unique permutations: [1,1,2], [1,2,1], and ...