Permutation Sequence @leetcode

转载自：http://blog.unieagle.net/2012/10/16/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Apermutation-sequence/

这道题其实有很强的规律可循。首先，n个元素的排列总数是n!。在下面的分析中，让k的范围是0 <= k < n!。（题目代码实际上是1<=k<=n!)

可以看到一个规律，就是这n！个排列中，第一位的元素总是(n-1)!一组出现的，也就说如果p = k / (n-1)!，那么排列的最开始一个元素一定是arr[p]。
这个规律可以类推下去，在剩余的n-1个元素中逐渐挑选出第二个，第三个，...，到第n个元素。程序就结束。

题目描述：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.

代码：20ms过大集合。

class Solution {
public:
    string getPermutation(int n, int k) {
        char *arr = new char[n];
        int pro = 1;
        for(int i = 0 ; i < n; ++i) {
            arr[i] = '1' + i;
            pro *= (i + 1);
        }
        k = k - 1;
        k %= pro;
        pro /= n;//pro = 1 * 2 * ... * (n-1)
        for(int i = 0 ; i < n-1; ++i) {
            //for index i
            int selectI = k / pro;
            k = k % pro;
            pro /= (n - i - 1);
            int temp = arr[selectI + i];
            for(int j = selectI; j > 0; --j)
                arr[i + j] = arr[i + j - 1];
            arr[i] = temp;
        }
        return string(arr, arr + n);
    }