这代码网络上随处可见，不过我分享的不只是递归，更重要的是分享 分析递归 的方法。/
排列、组合、子集都是常见的面试题，我希望能够帮助大家轻松的秒杀那些小菜鸟。
#include <stdio.h>
#include <string.h>

#define NUM 4
#define MAX_DEPTH 2

int mark[NUM+1] = {0};
int printList[NUM+1] = {0};

FILE* Log;

void bgOpenLog(void);

void bgWriteLog(const char* s);

void bgCloseLog(void);

void perm(int depth);

int main(void)
{
    bgOpenLog();

    perm(1);

    bgCloseLog();

    getchar();

    return 0;
}

void perm(int depth)
{
    char s[40] = {0};
    int i, j, k;

    for (i = 1; i <= NUM; i++)
    {
        if (!mark[i])
        {
            for (k = 0; k < depth; k++)
            {
                bgWriteLog("    ");
            }

            sprintf(s, "perm: (%d): %d", depth, i);
            bgWriteLog(s);
            bgWriteLog("\r\n");

            mark[i] = 1;
            printList[depth] = i;

            if (depth < MAX_DEPTH)
            {

                perm(depth+1);
            }
            else
            {
                for (j = 1; j <= MAX_DEPTH; j++)
                {
                    printf("%d ", printList[j]);
                }

                printf("\n");
            }

            mark[i] = 0;
        }
    }
}

void bgOpenLog(void)
{
    if(Log == NULL)
       Log = fopen("log.txt", "wb");
}

void bgWriteLog(const char* s)
{
    fwrite(s, sizeof(char), strlen(s), Log);
}

void bgCloseLog(void)
{
    fclose(Log);
    Log = NULL;
}

递归调用过程
    perm: (1): 1
        perm: (2): 2
        perm: (2): 3
        perm: (2): 4
    perm: (1): 2
        perm: (2): 1
        perm: (2): 3
        perm: (2): 4
    perm: (1): 3
        perm: (2): 1
        perm: (2): 2
        perm: (2): 4
    perm: (1): 4
        perm: (2): 1
        perm: (2): 2
        perm: (2): 3

[ 本帖最后由 BlueGuy 于 2011-1-7 22:50 编辑 ]

写一句话与大家共勉, "想写好程序，数据结构和算法一定得过关"