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. 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. 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…
# -*- 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…
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…
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…
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…
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…
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…
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…