Leetcode - Permutation Sequence

原题链接：https://leetcode.com/problems/permutation-sequence/
[分析]
思路1：调用 k 次NextPermutation.
思路2：数学解法，在n!排列中，每个第一位元素带领 (n-1)! 个排列数，假设 p = k / (n-1)!，则num[p]就是第一位上的数字，注意 k 要从0开始计数，因此进主循环前k--。
类似的方法求出剩下位置的数字。参考：
http://blog.csdn.net/doc_sgl/article/details/12840715
http://m.blog.csdn.net/blog/linhuanmars/22028697

public class Solution {
    public String getPermutation(int n, int k) {
        if (n <= 0 || n > 9)
            return "";
        int[] num = new int[n];
        for (int i = 0; i < n; i++)
            num[i] = i + 1;
        int factorial = 1;
        for (int i = 2; i <= n; i++)
            factorial *= i;
        StringBuilder result = new StringBuilder(n);
        k--;
        for (int i = 0; i < n; i++) {
            factorial /= n - i;
            int selectIdx = k / factorial;
            result.append(num[selectIdx]);
            k %= factorial;
            for (int j = selectIdx; j < n - i - 1; j++)
                num[j] = num[j + 1];
        }
        return result.toString();
    }
    public String getPermutation1(int n, int k) {
        if (n <= 0 || n > 9)
            return "";
        int[] num = new int[n];
        for (int i = 0; i < n; i++)
            num[i] = i + 1;
        for (int i = 1; i < k; i++)
            nextPermutation(num);
        StringBuilder result = new StringBuilder(n);
        for (int i = 0; i < n; i++)
            result.append(num[i]);
        return result.toString();
    }
    public void nextPermutation(int[] num) {
        int p = num.length - 2;
        while (p >= 0 && num[p] >= num[p + 1])
            p--;
        if (p >= 0) {
            int q = p + 1;
            while (q < num.length && num[q] > num[p])
                q++;
            int tmp = num[--q];
            num[q] = num[p];
            num[p] = tmp;
        }
        int q = num.length - 1;
        p = p + 1;
        while (p < q) {
            int tmp = num[p];
            num[p] = num[q];
            num[q] = tmp;
            p++;
            q--;
        }
    }
}