[LeetCode]Permutation Sequence

標籤：

題目描述:(連結)

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.

解題思路:

按照上一題的思路，暴力破解，逾時了，k可以超過n!，冏！代碼如下，程式碼摺疊功能，別開啟：

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());    }};

　　

[LeetCode]Permutation Sequence