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):

"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.

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

思路是对的，第1个数就是k/(n-1)!，k=k%(n-1)!，第二个数就是k/(n-2)!……

注意这类取余操作得到的是以0开始计数。一开始要把k--。

删除掉一个数之后，需要更新第i个数的值。这里用了一个数组，删掉一个数后，把后面的数往前移。

class Solution {
public:
    string getPermutation(int n, int k) {
        int a = 1;
        vector<int> kth(n); 
        
        for (int i = 1; i <= n - 1; ++i) {
            kth[i - 1] = i;
            a *= i;
        }
        kth[n - 1] = n;
        
        string ret = "";  
        n--; k--;
        while (n >= 1) {
            int v = k / a;                
            ret += '0' + kth[v];
            for (int i = v; i < n; ++i) kth[i]=kth[i+1];
            k = k % a;
            a = a / n;
            n--;
        }
        ret += '0' + kth[0];
        
        return ret;
    }
};

看了网上的代码，再重构了一下.

class Solution {
public:
    string getPermutation(int n, int k) {
        k--;
        int a = 1;
        vector<int> kth(n); 
        
        for (int i = 1; i <= n; ++i) {
            kth[i - 1] = i;
            a *= i;
        }
    
        string ret = "";  
        for (int i = 0; i < n; ++i) {
            a = a / (n - i);
            int v = k / a;
            k = k % a;
            ret.push_back(kth[v] + '0');
            for (int j = v; j < n - i - 1; ++j) kth[j] = kth[j + 1];
        }
        
        return ret;
    }
};