[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):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the k^th permutation sequence.

https://oj.leetcode.com/problems/permutation-sequence/

计算1~n数字的第k个排列

思路1：从小到大生成同时计数，直到第k个。 肯定超时试都不用试。。

思路2：直接根据规律计算第k个排列。

    分析：1~n个数共有n!个排列，1开头的有(n-1)!个，2开头的(n-1)!个，...n开头的有(n-1)!个。因此用k/(n-1)!就确定了第一位数字，然后依次类推（用过的数字要去除），继续在(n-1)!个数中找第k%(n-1)!个数。

思路2代码：

 

public class Solution {
    public String getPermutation(int n, int k) {
        int[] num = new int[n];
        int permSum = 1;
        for (int i = 0; i < n; i++) {
            num[i] = i + 1;
            permSum *= (i + 1);
        }
        StringBuilder sb = new StringBuilder();
        k--;//change to base 0
        for (int i = 0; i < n; i++) {
            permSum = permSum / (n - i);
            int selected = k / permSum;
            sb.append(num[selected]);
            for (int j = selected; j < n - i - 1; j++)
                num[j] = num[j + 1];
            k = k % permSum;
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        System.out.println(new Solution().getPermutation(4, 10));
    }

}

 

第二遍记录：

注意k每次的变化 k %=permProd;

注意每次选完一个元素后，数组后面的元素向前移动。移动之后被选择的数字被覆盖，数组最后一个元素舍弃。比如原来是[1,2,3,4,5]，3被选走了，数组变为[1,2,4,5,5]，最后一个5其实相当于被舍弃，之后只能从[1,2,4,5]中选下一次填充的数字。

public class Solution {
    public String getPermutation(int n, int k) {
        int[] num = new int[n];
        int permProd = 1;

        for (int i = 0; i < n; i++) {
            num[i] = i + 1;
            permProd *= num[i];
        }
        StringBuilder sb = new StringBuilder();
        k--;
        int idx = 0;
        for (int i = 0; i < n; i++) {
            permProd /= n - i;
            idx = k / permProd;
            sb.append(num[idx]);
            // move the array after the index idx
            for (int j = idx; j < n - i - 1; j++) {
                num[j] = num[j + 1];
            }
            k %= permProd;
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        System.out.println(new Solution().getPermutation(3, 6));
    }
}

 

 

参考：

http://blog.csdn.net/havenoidea/article/details/12837441

http://www.cnblogs.com/TenosDoIt/p/3721918.html