leetcode Permutations II

来源：互联网发布：微星主机知乎编辑：程序博客网时间：2024/06/14 05:54

先上题：

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].

这题的思路和Permutation一样，但是这里要去除重复的数字，我在这采用的方法是利用一个map，记录用过的数字，如果用过，则弃置不用。上代码（accept）：

class Solution {public:void dfs(vector<vector<int> >& result, vector<int>& tmp, vector<int>& num){if (num.empty()){result.push_back(tmp);return;}map<int, bool> mapping;for (int i = 0; i < num.size(); i++){mapping[num[i]] = false;}for (int i = 0; i < num.size(); i++){if (mapping[num[i]] == false){tmp.push_back(num[i]);mapping[num[i]] = true;auto iter = num.begin();iter += i;vector<int> tmpA = num;num.erase(iter);dfs(result, tmp, num);num = tmpA;tmp.pop_back();}}}vector<vector<int> > permuteUnique(vector<int> &num){vector<vector<int> > result;if (num.empty()){return result;}vector<int> tmp;dfs(result, tmp, num);return result;}};

0 0