使用java生成int数组的排列 - 错误 [英] Generating permutations of an int array using java -- error

问题描述

我正在编写一个JAVA代码来生成整数数组的所有排列。
虽然我得到的排列数正确，但排列本身并不正确。

I am writing a JAVA code to generate all permutations of a integer array. Though I am getting the number of permutations right, the permutations themselves are not correct.

在运行时我获得：

Input array Length
3
1
2
3
0Permutation is
1,  2,  3,  
##########################
1Permutation is
1,  3,  2,  
##########################
2Permutation is
3,  1,  2,  
##########################
3Permutation is
3,  2,  1,  
##########################
4Permutation is
1,  2,  3,  
##########################
5Permutation is
1,  3,  2,  
##########################
6  number of permutations obtained
BUILD SUCCESSFUL (total time: 3 seconds)


public class PermulteArray {

    public static int counter = 0;

    public static void Permute(int[] input, int startindex) {
        int size = input.length;

        if (size == startindex + 1) {
            System.out.println(counter + "Permutation is");
            for (int i = 0; i < size; i++) {
                System.out.print(input[i] + ",  ");
            }
            System.out.println();
            System.out.println("##########################");
            counter++;
        } else {
            for (int i = startindex; i < size; i++) {

                int temp = input[i];
                input[i] = input[startindex];
                input[startindex] = temp;
                Permute(input, startindex + 1);
            }
        }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("Input array Length");
        int arraylength = in.nextInt();
        int[] input = new int[arraylength];
        for (int i = 0; i < arraylength; i++) {
            input[i] = in.nextInt();
        }
        counter = 0;
        Permute(input, 0);
        System.out.println(counter + "  number of permutations obtained");
    }
}

推荐答案

int temp=input[i];
input[i]=input[startindex];
input[startindex]=temp;
Permute(input, startindex+1);

您在调用Permute之前交换了一个元素，但之后需要再次交换它以保持一致元素在for循环迭代中的位置。

You've swapped an element before calling Permute but you need to swap it back again afterwards to keep consistent positions of elements across iterations of the for-loop.

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