[LeetCode] Permutation Sequence

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the k^th permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

DFS从小到大枚举排列，到K时输出。

class Solution {
private:
    int a[10];
    bool canUse[10];
    string ret;
public:
    void dfs(int dep, int maxDep, int &k)
    {
        if (k == 0)
            return;
            
        if (dep == maxDep)
        {
            k--;
            if (k == 0)
            {
                ret = "";
                for(int i = 0; i < maxDep; i++)
                    ret += (char)(a[i] + '0');
                return;
            }
        }
        
        for(int i = 1; i <= maxDep; i++)
            if (canUse[i])
            {
                canUse[i] = false;
                a[dep] = i;
                dfs(dep + 1, maxDep, k);
                canUse[i] = true;
            }
    }
    
    string getPermutation(int n, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        memset(canUse, true, sizeof(canUse));
        
        dfs(0, n, k);
        
        return ret;
    }
};