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 [2,1,1].   先排序,如果一个元素与上一个元素相等,且前面没有使用该元素,则该元素不参与当前排…

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]. 求数组元素的全排列,数组不含重复元素 算法1:递归 类似于DFS的递归. 对于包含n个元素的数组,先确定第一位置的元素,…

Next Permutation: Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending ord…

题目: LeetCode46 I Given a collection of distinct numbers, return all possible permutations. (Medium) 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] ] LeetCode47 II Given a collection of numb…

作者:jostree  转载请注明出处 http://www.cnblogs.com/jostree/p/4051169.html 题目链接:leetcode Permutations II 无重全排列 题目要求对有重复的数组进行无重全排列.为了保证不重复,类似于全排列算法使用dfs,将第一个数字与后面第一次出现的数字交换即可.例如对于序列1,1,2,2 有如下过程: ,1,2,2 -> 1,,2,2 -> 1,1,,2 -> 1,1,2,  -> 1,2,,2 -> 1,2…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 47: Permutations IIhttps://oj.leetcode.com/problems/permutations-ii/ Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example,[1,1,2…

这是使用DFS来解数组类题的典型题目,像求子集,和为sum的k个数也是一个类型 解题步骤: 1:有哪些起点,例如,数组中的每个元素都有可能作为起点,那么用个for循环就可以了. 2:是否允许重复组合 3:处理某个数,判断结果 4:dfs递归 5:还原现场 一:Permutations Given a collection of numbers, return all possible permutations. For example,[1,2,3] have the following per…

字符串排列和PermutationsII差不多 Permutations第一种解法: 这种方法从0开始遍历,通过visited来存储是否被访问到,level代表每次已经存储了多少个数字 class Solution { public: vector<vector<int>> permute(vector<int>& nums) { vector<vector<int> > result; if(nums.empty()) return r…

目录 题目链接 注意点 解法 小结 题目链接 Permutations II - LeetCode 注意点 不确定有几种排列 解法 解法一:因为有重复的数字所以排列的个数不确定几个,一直生成新的排列直到和原始的数列相同为止 class Solution { public: vector<int> nextPermutation(vector<int> nums) { int n = nums.size(),i = n-2,j = n-1; while(i >= 0 &…

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 [2,1,1]. 首先分析一下与Permutations有何差异. 记当前位置为start,当前排列数…

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 [2,1,1]. SOLUTION 1: 还是经典的递归模板.需要处理的情况是:我们先把Num排序,然…

leetcode47. Permutations II 题意: 给定可能包含重复的数字的集合,返回所有可能的唯一排列. 思路: 这题全排列两种写法. 用hash表记录已经visit的数字,虽然看起来很朴实,但是比另一种方法稳多了,而且hash表的查找也是很高效的. 另一种用swap进行交换,确保已经遍历的点在前面,未遍历的点在后面.这个方法在没有duplicate的情况下,比如上一题,看不出有什么坑点.有了duplicate之后感觉很鸡儿皮. 首先要用sort排序,不然的话还是会遍历到重复的点的…

1. Permutations 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] ]  思路:直接使用递归来遍历即可,因为数字是不重复的所以,所以这里的每层递归都从索引0开始遍历整个数组…

Permutations 问题:给定一个无重复元素的数组,输出其中元素可能的所有排列 示例: 输入:[2,3,4] 输出:[ [2,3,4], [2,4,3], [3,2,4], [3,4,2], [4,2,3], [4,3,2] ] 解决思路:循环加递归 class Solution(object): def permute(self, nums): """ :type nums: List[int] :rtype: List[List[int]] ""…

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 [2,1,1]. 思路:这题相比于上一题,是去除了反复项. 代码上与上题略有区别.详细代码例如以…

Leetcode之回溯法专题-47. 全排列 II(Permutations II) 给定一个可包含重复数字的序列,返回所有不重复的全排列. 示例: 输入: [1,1,2] 输出: [ [1,1,2], [1,2,1], [2,1,1] ] 分析:跟46题一样,只不过加了一些限制(包含了重复数字). AC代码(时间复杂度比较高,日后二刷的时候改进): class Solution { List<List<Integer>> ans = new ArrayList<>()…

47.Permutations II (Medium)](https://leetcode.com/problems/permutations-ii/description/) [1,1,2] have the following unique permutations: [[1,1,2], [1,2,1], [2,1,1]] 题目描述:   数组元素可能包含重复元素,给出其数组元素的所有排列,不包含重复的排列. 思路分析:   在实现上,和Permutations不同的是要先排序,然后在添加元…

题目要求: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 [2,1,1]. 代码如下: class Solution { public: vector…

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 [2,1,1]. 这道题是之前那道Permutations 全排列的延伸,由于输入数组有可能出现重复数字,如果按照之前的算法运算,会有…

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 [2,1,1]. 思路:有重复数字的情况,之前在Subsets II,我们采取的是在某一个递归内,用for循环处理所有重复数字.这里当…

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 [2,1,1]. 解题思路: Permutations 那题输入是不包含重复元素的,故生成的排列都是不同的,II中输入元素可能相同 因…

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] ] 这道题是之前那道 Permutations 的延伸,由于输入数组有可能出现重复数字,如果按照之前的算法运算,会有重复排列产生,我们要避免重复的产生,在递归函数中要判断前面一…

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] ] 有重复数字的情况,之前在Subsets II,我们采取的是在某一个递归内,用for循环处理所有重复数字.这里也相同,需要在递归内考虑重复数字,重复数字只能插入在已插入的重复…

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], [2,1,1] ] 分析: 全组合的思想,保证start和end之间交换的时候中间没有与end相同的数字 class Solution…

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], [2,1,1] ]和一般的Permutation不一样的是,这种permutation需要排序,使相同的元素能够相邻,选取下一个元素的时…

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 [2,1,1]. Solution: public class Solution { public List<List<Intege…

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], [2,1,1] ] 思路1.这题与Permutations的区别在于他允许重复数字,最简单的就是利用1的结果,保存在一个set中,去重复…

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 [2,1,1]. 解题思路一: 发现Java for LeetCode 046 Permutations自己想多了,代码直接拿来用…

题目: 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 [2,1,1]. 代码: class Solution { public: vector<vector<int>>…

题目 带重复元素的排列 给出一个具有重复数字的列表,找出列表所有不同的排列. 样例 给出列表 [1,2,2],不同的排列有: [ [1,2,2], [2,1,2], [2,2,1] ] 挑战 使用递归和非递归分别完成该题. 解题 和上面差不多,增加判断res中是否已经存在该排列的语句,这种方法不是很好,但是竟然也可以通过 class Solution { /** * @param nums: A list of integers. * @return: A list of unique perm…