字符组合排序

1.将给定字符组合后，按字典序排序输出，示例：
输入：a, b, c;
输出：abc，acb，bac，bca，cab，cba
2.将给定字符组合后，按字典序排序输出，示例：
输入：a, b, c;
输出：a，b，c，ab，ac，bc, abc

//Example 1
#include <stdio.h>
#include <string.h>

void combchar(char* str, int from, int to)
{
    if(from >= to){
        for (int i = 0; i <= to; ++i){
            PRINT("%c", str[i]);
        }
        PRINT(" ");
        return;
    }

    int j = 0;
    for(int i = from; i <= to; i++){
        char tmp = str[i];
        for(j = i; j-1 >= from; --j){
            str[j] = str[j-1];
        }
        str[j] = tmp;

        combchar(str, from+1, to);

        for(j = from; j+1 <= i; ++j){
            str[j] = str[j+1];
        }
        str[j] = tmp;

        if(from == 0) PRINT("\n");
    }
}

int main(){
    char str[] = {'a', 'b', 'c', 'd'};
    combchar(str, 0, sizeof(str)-1);
    return 0;
}

Example 1 Result 结果如下所示：

Example 1 Result:
abcd abdc acbd acdb adbc adcb 
bacd badc bcad bcda bdac bdca 
cabd cadb cbad cbda cdab cdba 
dabc dacb dbac dbca dcab dcba

//Example 2
#include <stdio.h>
#include <string.h>
#define PRINT   printf
int num = 0;
void combination(char* str, int from, int to, int len, int t)
{
    if(len == 1){
        PRINT("%c\n", str[from]); ++num;
        for(int j = from+1; j <= to; j++){
            PRINT("%*c\n", 3*t+1, str[j]);
            ++num;
        }
        return;
    }
    for(int j = from; j <= to-len+1; j++){//1,2,3,4
        PRINT("%*c->", (j==from) ? 1 : 3*t+1, str[j]);
        combination(str, j+1, to, len-1, t+1);
    }
}

void combination2(char* str, int from, int to, int len, int t, char* tmpStr)
{
    if(len == 1){
        for(int j = from; j <= to; j++){
            tmpStr[t] = str[j];
            for(int i = 0; i <= t; i++){
                PRINT("%c", tmpStr[i]);
            }
            PRINT(" ");
        }
    }else{
        for(int j = from; j <= to-len+1; j++){//1,2,3,4
            tmpStr[t] = str[j];
            combination2(str, j+1, to, len-1, t+1, tmpStr);

        }
    }
    if(t == 0) PRINT("\n");
}

void findSubStr(char* str, int from, int to)
{
    char tmpStr[from-to+1];
    for(int j = 1; j <= to-from+1; j++){//len=1,2,3,4
        //combination(str, from, to, j, 0);
        combination2(str, j+1, to, len-1, t+1, tmpStr);
        PRINT("-------------------%d|%d\n", j, num);
    }
}

int main(){
    char str[] = {'a', 'b', 'c', 'd'};
    int size = strlen(str);
    findSubStr(str, 0, size-1);
    return 0;
}

Example 2 Result 结果如下所示：

Example 2 combination Result:
a
b
c
d
a->b
   c
   d
b->c
   d
c->d
a->b->c
      d
   c->d
b->c->d
a->b->c->d

Example 2 combination2 Result
a b c d 
ab ac ad bc bd cd 
abc abd acd bcd 
abcd