[leetcode] Permutations II

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：

含有重复数字的全排列，关键是去重。首先排序，重复的元素相邻。如果当前元素与之前元素相等，如果之前元素还未被访问，则当前元素不可访问。换句话说，当前元素与前一个元素相等，只有当前一个元素被访问时，当前元素才能被访问。从而实现了去重。

题解：

class Solution {
public:
    vector<vector<int> > res;
    vector<int> tmp;
    void dfs(vector<int> &num, bool visited[], int index) {
        if(index==num.size()) {
            res.push_back(tmp);
            return;
        }
        for(int i=0;i<num.size();i++) {
            if(!visited[i]) {
                if(i!=0 && num[i]==num[i-1] && !visited[i-1])
                    continue;
                tmp.push_back(num[i]);
                visited[i] = true;
                dfs(num, visited, index+1);
                visited[i] = false;
                tmp.pop_back();
            }
        }
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
        int n = num.size();
        bool *visited = new bool [n];
        memset(visited, false, sizeof(bool)*n);
        sort(num.begin(), num.end());
        dfs(num, visited, 0);
        return res;
    }
};