60. 排列序列

给出集合 [1,2,3,...,n]，其所有元素共有 n! 种排列。

按大小顺序列出所有排列情况，并一一标记，当 n = 3 时, 所有排列如下：

"123"
"132"
"213"
"231"
"312"
"321"
给定 n 和 k，返回第 k 个排列。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/permutation-sequence
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原问题

import java.util.Arrays;

class Solution {
    public String getPermutation(int n, int k) {
        int[] factorial = new int[n];
        factorial[0] = 1;
        for (int i = 1; i < n; ++i) {
            factorial[i] = factorial[i - 1] * i;
        }

        StringBuilder ans = new StringBuilder();
        int[] valid = new int[n + 1];
        Arrays.fill(valid, 1);
        for (int i = 1; i <= n; ++i) {
            int order = (k - 1) / factorial[n - i] + 1;
            for (int j = 1; j <= n; ++j) {
                order -= valid[j];
                if (order == 0) {
                    ans.append(j);
                    valid[j] = 0;
                    break;
                }
            }
            k = (k - 1) % factorial[n - i] + 1;
        }
        return ans.toString();
    }
}

官方题解思考题

iimport java.util.Arrays;
import java.util.Scanner;

class Solution {

    private int getK(String str) {
        int n = str.length();
        int s = 1;
        for (int i = 2; i <= n; ++i) {
            s *= i;
        }

        int[] valid = new int[n + 1];
        Arrays.fill(valid, 1);

        int ret = 0;
        for (int i = 0; i < str.length(); ++i) {
            s /= (n--);
            int digit = str.charAt(i) - '0';
            int r = 0;
            for (int j = 1; j <= digit; ++j) {
                r += valid[j];
            }
            ret += (r - 1) * s;
            valid[digit] = 0;
        }

        return ret + 1;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            System.out.println(new Solution().getK(in.next()));
        }
    }
}