47. Permutations II (java )

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] ] 1 public class Solution { 2 List<List<Integer>> ans

16. Permutations II/47. Permutations II 本题难度: Medium Topic: Search & Recursion Description Given a list of numbers with duplicate number in it. Find all unique permutations. Example Example 1: Input: [1,1] Output: [ [1,1]] Example 2: Input: [1,2,2] O

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.详细代码例如以下

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/ 4/15

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 used随着计

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]. public class Solution { List<Integer> seq; List<List&l

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. 1 class Solution(object): 2 def __ini

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