打印出阵列的所有排列

Question

我正在研究一个程序,我有一个函数可以交换用户输入的长度数组中的位置.但是,我想弄清楚如何打印出这个函数调用N!times,列出函数中的所有排列.

我的置换函数代码是:

static void nextPerm(int[] A){
    for( int i = (n-1); i > 0; i-- ){
        if( A[i] < A[i+1] ){
            A[i] = pivot;
            continue;
        }
        if( A[i] >= A[i+1] ){
            reverseArray(A);
            return;
        }
    }

    for( int i = n; i > 0; i--){
        if( A[i] > pivot ){
            A[i] = successor;
            continue;
        }
    }

    Swap(pivot, successor);

    int[] B = new int[pivot+1];
    reverseArray(B);

    return;
}

我应该在函数main中编写一个循环,这将打印出来!次？

Answer 1

与纯粹迭代相比,作为递归和迭代的组合,创建(或打印)数组的排列更容易.肯定有迭代方法可以做到这一点,但组合起来却特别简单.具体来说,请注意,根据定义N!长度为N阵列的排列 - 第一个槽的N个选择,第二个槽的N-1个选择等等.因此,我们可以将算法分解为数组中每个索引i的两个步骤.

1. 选择子数组arr[i....end]中的ith元素作为数组的元素.将元素与当前元素交换arr[i].
2. 递归置换arr[i+1...end].

我们注意到,这将在O(N!)运行,作为第一个呼叫N个子通话将被制成,其中的每一个将N-1次电话,等等等等.此外,每一个元素最终会在每个位置上是,只要只进行交换,就不会复制任何元素.

public static void permute(int[] arr){
    permuteHelper(arr, 0);
}

private static void permuteHelper(int[] arr, int index){
    if(index >= arr.length - 1){ //If we are at the last element - nothing left to permute
        //System.out.println(Arrays.toString(arr));
        //Print the array
        System.out.print("[");
        for(int i = 0; i < arr.length - 1; i++){
            System.out.print(arr[i] + ", ");
        }
        if(arr.length > 0) 
            System.out.print(arr[arr.length - 1]);
        System.out.println("]");
        return;
    }

    for(int i = index; i < arr.length; i++){ //For each index in the sub array arr[index...end]

        //Swap the elements at indices index and i
        int t = arr[index];
        arr[index] = arr[i];
        arr[i] = t;

        //Recurse on the sub array arr[index+1...end]
        permuteHelper(arr, index+1);

        //Swap the elements back
        t = arr[index];
        arr[index] = arr[i];
        arr[i] = t;
    }
}

样本输入,输出:

public static void main(String[] args) {
    permute(new int[]{1,2,3,4});
}

[1, 2, 3, 4]
[1, 2, 4, 3]
[1, 3, 2, 4]
[1, 3, 4, 2]
[1, 4, 3, 2]
[1, 4, 2, 3]
[2, 1, 3, 4]
[2, 1, 4, 3]
[2, 3, 1, 4]
[2, 3, 4, 1]
[2, 4, 3, 1]
[2, 4, 1, 3]
[3, 2, 1, 4]
[3, 2, 4, 1]
[3, 1, 2, 4]
[3, 1, 4, 2]
[3, 4, 1, 2]
[3, 4, 2, 1]
[4, 2, 3, 1]
[4, 2, 1, 3]
[4, 3, 2, 1]
[4, 3, 1, 2]
[4, 1, 3, 2]
[4, 1, 2, 3]

Answer 2

我大部分时间都遵循这种方法..(由Robert Sedgewick和Kevin Wayne提供).

public class Permutations {

    // print N! permutation of the characters of the string s (in order)
    public  static void perm1(String s) { perm1("", s); }
    private static void perm1(String prefix, String s) {
        int N = s.length();
        if (N == 0) System.out.println(prefix);
        else {
            for (int i = 0; i < N; i++)
               perm1(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, N));
        }

    }

    // print N! permutation of the elements of array a (not in order)
    public static void perm2(String s) {
       int N = s.length();
       char[] a = new char[N];
       for (int i = 0; i < N; i++)
           a[i] = s.charAt(i);
       perm2(a, N);
    }

    private static void perm2(char[] a, int n) {
        if (n == 1) {
            System.out.println(a);
            return;
        }
        for (int i = 0; i < n; i++) {
            swap(a, i, n-1);
            perm2(a, n-1);
            swap(a, i, n-1);
        }
    }  

    // swap the characters at indices i and j
    private static void swap(char[] a, int i, int j) {
        char c;
        c = a[i]; a[i] = a[j]; a[j] = c;
    }

但是,还有一种更简单的方法可以做到这一点.也许你可以在这周围工作

class PermutingArray {
    static void permutingArray(java.util.List<Integer> arrayList, int element) {
        for (int i = element; i < arrayList.size(); i++) {
            java.util.Collections.swap(arrayList, i, element);
            permutingArray(arrayList, element + 1);
            java.util.Collections.swap(arrayList, element, i);
        }
        if (element == arrayList.size() - 1) {
            System.out.println(java.util.Arrays.toString(arrayList.toArray()));
        }
    }

    public static void main(String[] args) {
        PermutingArray
                .permutingArray(java.util.Arrays.asList(9, 8, 7, 6, 4), 0);
    }
}

这里的工作示例.. IDeone链接