LeetCode: Permutation Sequence 解题报告

Permutation Sequence

    https://oj.leetcode.com/problems/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.

Note: Given n will be between 1 and 9 inclusive.

解答：

1. 以某一数字开头的排列有(n-1)! 个。
例如： 123， 132， 以1开头的是 2！个
2. 所以第一位数字就可以用 （k-1） / (n-1)!  来确定 .这里K-1的原因是，序列号我们应从0开始计算，否则在边界时无法计算。
3. 第二位数字。假设前面取余后为m，则第二位数字是 第 m/(n-2)! 个未使用的数字。
4. 不断重复2，3，取余并且对(n-k)!进行除法，直至计算完毕

以下为主页君的代码，敬请指正：

解法1：

采用比较复杂的boolean来计算数字的索引（我们需要用一个boolean的数组来记录未使用的数字）：

package Algorithms.permutation;

/*
 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.
 * */
public class PermutationSequence {
    public static String getPermutation(int n, int k) {
        if (n == 0) {
            return "";
        }
        
        // 先计算出(n)!
        int num = 1;
        for (int i = 1; i <= n; i++) {
            num *= i;
        }
        
        boolean[] use = new boolean[n];
        for (int i = 0; i < n; i++) {
            use[i] = false;
        }
        
        // 因为index是从0开始计算 
        k--;
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < n; i++) {
            // 计算完第一个数字前，num要除以(n)
            num = num / (n - i);
            
            int index = k / num;
            k = k % num;
            
            for (int j = 0; j < n; j++) {                
                if (!use[j]) {
                    if (index == 0) {
                        // 记录下本次的结果.
                        sb.append((j + 1) + "");
                        use[j] = true;
                        break;
                    }
                    
                    // 遇到未使用过的数字，记录index
                    index--;
                }
            }
        }
        
        return sb.toString();
    }

    public static void main(String[] args) {
        System.out.println(getPermutation(3, 5));
    }

}

 

解法2：

优化后，使用链表来记录未使用的数字，每用掉一个，将它从链表中移除即可。

public String getPermutation1(int n, int k) {
        // 1:17 -> 1:43
        LinkedList<Character> digits = new LinkedList<Character>();
        
        // bug 2: should only add n elements.
        for (char i = '1'; i <= '0' + n; i++) {
            digits.add(i);
        }
        
        k = k - 1;
        StringBuilder sb = new StringBuilder();
        
        int sum = 1;
        // n!
        for (int i = 1; i <= n; i++) {
            sum *= i;
        }
        
        int cur = n;
        while (!digits.isEmpty()) {
            sum /= cur;
            cur--;
            
            int digitIndex = k / sum;
            k = k % sum;
            //Line 25: error: cannot find symbol: method digits(int)
            sb.append(digits.get(digitIndex));
            // remove the used digit.
            digits.remove(digitIndex);
        }
        
        return sb.toString();
    }

 

解法3：

在2解基础进一步优化，使用for 循环替代while 循环，更简洁：

public String getPermutation(int n, int k) {
        // 1:17 -> 1:43
        LinkedList<Character> digits = new LinkedList<Character>();
        
        // bug 2: should only add n elements.
        for (char i = '1'; i <= '0' + n; i++) {
            digits.add(i);
        }
        
        // The index start from 0;
        k--;
        StringBuilder sb = new StringBuilder();
        
        int sum = 1;
        // n!
        for (int i = 1; i <= n; i++) {
            sum *= i;
        }
        
        for (int i = n; i >= 1; i--) {
            sum /= i;
            int digitIndex = k / sum;
            k = k % sum;
            
            //Line 25: error: cannot find symbol: method digits(int)
            sb.append(digits.get(digitIndex));
            
            // remove the used digit.
            digits.remove(digitIndex);
        }
        
        return sb.toString();
    }

GitHub代码链接