[LeetCode]Permutation Sequence

题目描述:(链接)

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 kth permutation sequence.

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

解题思路:

按照上一题的思路，暴力破解，超时了，k可以超过n!，冏！代码如下，代码折叠，别打开：

class Solution {
public:
    string getPermutation(int n, int k) {
        string result(n, '0');
        for (size_t i = 0; i < result.size(); ++i) {
            result[i] += i + 1;
        }
        
        for (size_t i = 0; i < k; ++i) {
            nextPermutation(result);
        }
        
        return result;
    }
    
    void nextPermutation(string& str) {
        int i;
        int j;
        // From right to left, find the first item(PartitionNumber) which violates the increase trend
        for (i = str.size() - 2; i >= 0; --i) {
            if (str[i] < str[i + 1]) break;
        }
        
        // From right to left, find the first item(ChangeNumber) which is larger than PartitionNumber
        for (j  = str.size(); j >= i; --j) {
            if (str[j] > str[i]) break;
        }
        
        // swap PartitionNumber and ChangeNumber
        if (i >= 0) {
            swap(str[i], str[j]);
        }
        
        // reverse all after PartitionNumber index
        reverse(str.begin() + i + 1, str.end());
    }
};

转载:http://www.cnblogs.com/tenosdoit/p/3721918.html

有没有什么方法不是逐个求，而是直接构造出第k个排列呢？我们以n = 4，k = 17为例，数组src = [1,2,3,...,n]。

第17个排列的第一个数是什么呢：我们知道以某个数固定开头的排列个数 = (n-1)! = 3! = 6, 即以1和2开头的排列总共6*2 = 12个，12 < 17, 因此第17个排列的第一个数不可能是1或者2，6*3 > 17, 因此第17个排列的第一个数是3。即第17个排列的第一个数是原数组（原数组递增有序）的第m = upper(17/6) = 3（upper表示向上取整）个数。    

class Solution {
public:
    string getPermutation(int n, int k) {
        int total = factorial(n);
        string candidate = string("123456789").substr(0, n);
        string result(n, '0');
        for (int i =0; i < n; ++i) {
            total /= (n - i);
            int index = (k - 1) / total;
            result[i] = candidate[index];
            candidate.erase(index, 1);
            k -= index * total;
        }
        
        return result;
    }
private:
    int factorial(int n) {
       int result = 1;
       for (int i = 2; i <= n; ++i) {
           result *= i;
       }
       
       return result;
    }
};