生成下长度为N C语言中的所有字符串 [英] Generate all strings under length N in C

问题描述

我想编码这个自己和可怕的失败。这基本上是我想要的：

I tried coding this myself and horribly failed. This is basically what I want:

a
b
...
z
aa
ba
...
za
ab
bb
...
zz
aaa
baa
...
zzz

在结束它应该产生的每个字符串，它是那么短的N个字符与字符集AZ。所以我不是在寻找的排列（其中1001实现可以在互联网上找到），但对于组合，带替换（至少这是它被称为在Python） 。 顺序并不重要，速度。

In the end it should have generated every string that is shorter then N characters with charset a-z. So I'm not looking for permutations (of which 1001 implementations can be found on the internet), but for combinations with replacement (at least that's how it's called in Python). Order is not important, speed is.

推荐答案

看起来像你想在C，这里是一个办法做到这一点：

Looks like you want it in C, here is a way to do it:

#include <stdlib.h>
#include <stdio.h>

int inc(char *c){
    if(c[0]==0) return 0;
    if(c[0]=='z'){
        c[0]='a';
        return inc(c+sizeof(char));
    }   
    c[0]++;
    return 1;
}

int main(void){
    int n = 3;
    int i,j;
    char *c = malloc((n+1)*sizeof(char));
    for(i=1;i<=n;i++){
        for(j=0;j<i;j++) c[j]='a';
        c[i]=0;
        do {
            printf("%s\n",c);
        } while(inc(c));
    }
    free(c);
}

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