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], [2,1,1] ] 解答 C语言的实现以前总结过,请见 http://www.cnblogs.com/YuNanlong/p/6…

# -*- 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…

一天一道LeetCode系列 (一)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]. (二)解题 求全排列数.具体思路可以参考[一天一道LeetCode]#…

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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:递归 方法二:回溯法 日期 题目地址:https://leetcode.com/problems/permutations-ii/description/ 题目描述 Given a collection of numbers that might contain duplicates, return all possible unique p…

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]. 依旧是dfs,不过这次要注意的是排列不允许有重复的vector存在,那么在排列之前应该先排序,然后每次检查的时候,如…

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…

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…

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…

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] ] 46. Permutations 的拓展,这题数组含有重复的元素.解法和46题,主要是多出处理重复的数字. 先对nu…

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…

题目链接: https://leetcode.com/problems/permutations-ii/?tab=Description   给出数组,数组中的元素可能有重复,求出所有的全排列   使用递归算法:   传递参数 List<List<Integer>> list, List<Integer> tempList, int[] nums, boolean[] used   其中list保存最终结果 tempList保存其中一个全排列组合 nums保存初始的数组…

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]. 链接: http://leetcode.com/problems/permutations-ii/ 题解:…

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. 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] ] 本题有重复元素,条件较难思考,做这个题费了小劲． 总体框架是应用 backtrack 解决． 样例:[1 1 2]…

思路是在相似题Permutations的基础上,将结果放到set中,利用set容器不会出现重复元素的特性,得到所需结果 但是利用代码中的/* */部分通过迭代器遍历set将set中的元素放在一个新的vector中时,会出现memory limit exceeded错误(原因??) 上网查找后发现可以直接通过return vector<vector<int>> (mySet.begin(),mySet.end())得到结果,并且代码通过. class Solution { publi…

▶ 问题:字典序生成有关的问题. ▶ 31. 由当前序列生成字典序里的下一个序列. ● 初版代码,19 ms class Solution { public: void nextPermutation(vector<int>& nums) { int i, j, temp; ; i > && nums[i - ] >= nums[i]; i--);// 从右向左寻找第一个递增对 if (!i) // i == 0,当前排序已经是最大的,全反序得到最小的 {…

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] ] 与上一题不同,就是在19行加个判断即可. class Solution(object): def __init__(…

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].题解:依旧使用的是DFS的思想. 首先需要遍历输入数组,获取一共有多少种不同的数字,每个数字有多少个. 最简单的方法,…

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]. 排列组合的问题,不用考虑是否发生重复的情况,代码如下: class Solution { public: vector<vector<int>…

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], [2,1,1] ] 题解: Solution 1 (TLE) class Solution { public: void dfs…

Basic Calculator II 题目 思路 和这个一样:Basic Calculator I 代码 class ExpressionTransformation { public: string trans_to_postfix_expression_to_s(string); // 将得到的表达式转化为后缀表达式 long long int calculate_from_postfix_expression(); // 依据后缀表达式计算值 private: vector<string…

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] ] Time: O(N!)Space: O(N) class Solution: def permuteUnique(self, nums: List[int]) -> Lis…

Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. 例子: If nums = [1,2,2], a solution is: […

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form. For example: Given s = "aabb", return ["abba", "baab"]. Given s = "a…

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…

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<>()…

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…