怎样输出可重集的全排列

 #返回上一级

@Author: 张海拔

@Update: 2014-02-03

@Link: http://www.cnblogs.com/zhanghaiba/p/3533614.html

permutation1()可以输出[1, n]的全排列，通过实现这个函数回顾了一下回溯法。

比求n的全排列还基础的回溯法典型问题是：“怎样输出二叉树的所有路径（按字典序）？ link(public)”

permutation2()可以输出不可重集的全排列，主要是为了引出下一个函数。

permutation()可以输出可重复集合的全排列，功能完整。

三个函数均是独立可复用的，由于这个算法效率是指数级的，所以保存路径的path数组和vis标记数组可以放在递归函数中，同时设为static。

/*
 *Author: ZhangHaiba
 *Date: 2014-1-25
 *File: full_permutation.c
 *
 *a demo shows how to print full permutaion of [1,n] and Repeatable Set(a sorted string)
 */

#include <stdio.h>
#include <string.h>
#define LEN 128

//show permutation of [1, n]
void permutation1(int step, int n);

//guarantee sorted string and have not repeating characters
void permutation2(int step, char *sorted_string, int string_len);

//guarantee sorted string
void permutation(int step, char *sorted_string, int string_len);

int main(void)
{
    int n;
    char str[LEN];

    scanf("%d", &n);
    permutation1(0, n);
    scanf("%s", str);
    permutation2(0, str, strlen(str));
    scanf("%s", str);
    permutation(0, str, strlen(str));
    return 0;
}

void permutation1(int step, int n)
{
    static int path[LEN];
    static int vis[LEN];
    int i;

    if (step == n) {
        for (i = 0; i < n; ++i)
            printf(i == n-1 ? "%d
" : "%d", path[i]);
        return;
    }
    for (i = 0; i < n; ++i) {
        if (!vis[i]) {
            path[step] = i+1;
            vis[i] = 1;
            permutation1(step+1, n);
            vis[i] = 0;
        }
    }
}

void permutation2(int step, char *s, int n)
{
    static char path[LEN];
    static int vis[LEN];
    char *p;

    if (step == n) {
        path[n] = '';
        printf("%s
", path);
        return;
    }
    for (p = s; *p; ++p) {
        if (!vis[*p]) {
            path[step] = *p;
            vis[*p] = 1;
            permutation2(step+1, s, n);
            vis[*p] = 0;
        }
    }
}

void permutation(int step, char *s, int n)
{
    static char path[LEN];
    char *p;
    int i, cnt1, cnt2;

    if (step == n) {
        path[n] = '';
        printf("%s
", path);
        return;
    }
    for (p = s; *p; ++p) {
        if ( p == s || *p != *(p-1) ) { //for each repeatless character
            for (cnt1 = i = 0; i < step; ++i)
                if (*p == path[i])
                    cnt1++; //count *p in path
            for (cnt2 = i = 0; i < n; ++i)
                if (*p == s[i])
                    cnt2++; //count *p in source string
            if (cnt1 < cnt2) {
                path[step] = *p;
                permutation(step+1, s, n);
            }
        }
    }
}