[Leetcode 101] 47 Permutations II

Problem:

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].

Analysis:

An normal way to solve the problem is first generate all the possible permutations, and then compare it with all the existing permutations. If it's unique, then put it into the result vector. But it's too time consuming. So we should use another method.

The method is as follows: sort the given num vector. And each time we pick a new element, if the element is duplicated, then this element cannot be picked unless it previous same element has been picked. This strategy can help us skip those situations that might cause duplication problems.

Code:

class Solution {
public:
    vector<vector<int> > res;
    vector<int> t_res;
    int len;
    int *sig;

    vector<vector<int> > permuteUnique(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        res.clear();
        t_res.clear();
        
        len = num.size();
        sig = new int[len];
        for (int i=0; i<len; i++)
            sig[i] = 0;
        
        sort(num.begin(), num.end());
        
        dfs(0, num);
        
        return res;
    }
    
private:
    void dfs(int l, vector<int> &num) {
        if (l == len) {
            res.push_back(t_res);
            return ;
        }
        
        for (int i=0; i<len; i++) {
            if (sig[i] == 0 && isValid(i, num)) {
                t_res.push_back(num[i]);
                sig[i] = 1;
                dfs(l+1, num);
                sig[i] = 0;
                t_res.pop_back();
            }
        }
    }
    
    bool isValid(int i, vector<int> &num) {
        return (i==0) || (num[i-1] != num[i])|| ((num[i-1] == num[i]) && sig[i-1]);
    }
};