[Leetcode] 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].

 

注意不能用n！来判断了，可以先排一下序，当序列逆序时结束。

class Solution {
public:
    bool nextPermutation(vector<int> &num) {
        if (num.size() < 2) {
            return false;
        }
        int i, k;
        bool flag = true;
        for (i = num.size() - 2; i >= 0; --i) {
            if (num[i] < num[i + 1]) {
                break;
            }
        }
        if (i < 0) {
            flag = false;
        }
        for (k = num.size() - 1; k > i; --k) {
            if (num[k] > num[i]) {
                break;
            }
        }
        swap(num[i], num[k]);
        reverse(num.begin() + i + 1, num.end());
        return flag;
    }
    
    vector<vector<int> > permuteUnique(vector<int> &num) {
        vector<vector<int> > res;
        sort(num.begin(), num.end());
        res.push_back(num);
        while (nextPermutation(num)) {
            res.push_back(num);
        }
        return res;
    }
};