47. 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],  [2,1,1]]思路：    按照寻找下一序列的思路，先升序排序，然后依次找出逐渐增大的组合序列。类似DFS，深度优先遍历，先固定第一个元素，然后排列n-1个元素，采用递归方法(注意跳过重复的排列)。采用交换的方法来排列组合。*/#include <stdio.h>#include <iostream>#include <string>#include <sstream>#include <vector>#include <algorithm>using namespace std;class Solution {public: vector<vector<int>> permute2(vector<int>& nums) {        vector<vector<int> >res;sort(nums.begin(),nums.end());//升序dfs2(res,nums,0);return res; }private:    //nums 通过引用形式，，每次交换完元素后才交换回去，保证局部递增void dfs2(vector<vector<int> > &res,vector<int> &nums,int begin){if(begin==nums.size()-1){res.push_back(nums);return;}for(int i=begin;i<nums.size();i++)//寻找下一个递增序列组合{if(i!=begin && nums[i]==nums[begin]) continue;//跳过重复值swap(nums[begin],nums[i]);dfs2(res,nums,begin+1);swap(nums[begin],nums[i]);}}    //不通过引用形式void dfs(vector<vector<int> > &res,vector<int> nums,int begin){if(begin==nums.size()-1){res.push_back(nums);return;}for(int i=begin;i<nums.size();i++)//寻找下一个递增序列组合{if(i!=begin && nums[i]==nums[begin]) continue;//跳过重复值swap(nums[begin],nums[i]);dfs2(res,nums,begin+1);}}};int main(){    Solution mys;vector<std::vector<int> > res;vector<int> nums={1,2,3};res=mys.permute2(nums);for(int i=0;i<res.size();i++){    for(int j=0;j<res[i].size();j++)    cout<<res[i][j];cout<<endl;}    return 0;}