字符串数组的排列

Question

我正在解决这个挑战： https://www.hackerrank.com/challenges/permutations-of-strings/problem

我正在使用这个算法： https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order

假设我输入的字符串数组为{"a", "b", "c"}，那么输出应该是：

a b c 
a c b
b c a
c b a
b a c
c a b

因为有 3 个不同的字符串，所以有 3 个！ = 6 个排列。

程序也是要处理重复的，所以如果我输入{"a", "b", "b"}，就只有3个！ /2！ = 3 个排列。

无论如何，当我的程序到达c b a 时，它就会退出。为什么？我该如何解决？

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void swap(char** s1, char** s2) // to swap two strings
{
    char* temp = *s1;
    *s1 = *s2;
    *s2 = temp;
}
int next_permutation(int n, char **s)
{
    /**
    * Complete this method
    * Return 0 when there is no next permutation and 1 otherwise
    * Modify array s to its next permutation
    */
    int k, i, l;
    k = i = l = 0;

    for(i = n - 2; i >= 0; i--) // first step
        if(strcmp(s[i], s[i + 1]) < 0) {
            k = i;
            break;
        }


    if(i == -1)
        return 0;

    for(i = n - 1; i >= k + 1; i--) // second step
        if(strcmp(s[k], s[i]) < 0) {
            l = i;
            break;
        }


    swap(&s[k], &s[l]); // third step

    for(i = k + 1; i < n / 2; i++) // fourth step
        swap(&s[i], &s[n - i - 1]);

    return 1;
}

int main() // locked code
{
    char **s;
    int n;
    scanf("%d", &n);
    s = calloc(n, sizeof(char*));
    for (int i = 0; i < n; i++)
    {
        s[i] = calloc(11, sizeof(char));
        scanf("%s", s[i]);
    }
    do
    {
        for (int i = 0; i < n; i++)
            printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
    } while (next_permutation(n, s));
    for (int i = 0; i < n; i++)
        free(s[i]);
    free(s);
    return 0;
}

Answer 1

这是最终代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void swap(char** s1, char** s2) // swaps two strings
{
    char* temp = *s1;
    *s1 = *s2;
    *s2 = temp;
}
int next_permutation(int n, char **s) // function to complete
{
    int k, i, l;
    k = i = l = 0;

    for(i = n - 2; i >= 0; i--) // step 1
        if(strcmp(s[i], s[i + 1]) < 0) {
            k = i;
            break;
        }


    if(i == -1)
        return 0;

    for(i = n - 1; i >= k + 1; i--) // step 2
        if(strcmp(s[k], s[i]) < 0) {
            l = i;
            break;
        }


    swap(&s[k], &s[l]); // step 3

    int inner = k + 1 + n - 1; // step 4
    int cond = inner / 2;
    for(i = n - 1; i > cond; i--)
        swap(&s[i], &s[inner - i]);

    return 1;
}

int main() // locked code
{
    char **s;
    int n;
    scanf("%d", &n);
    s = calloc(n, sizeof(char*));
    for (int i = 0; i < n; i++)
    {
        s[i] = calloc(11, sizeof(char));
        scanf("%s", s[i]);
    }
    do
    {
        for (int i = 0; i < n; i++)
            printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
    } while (next_permutation(n, s));
    for (int i = 0; i < n; i++)
        free(s[i]);
    free(s);
    return 0;
}