题目28扩展：求字符的所有组合
eg:abc对应的组合a,b,c;a,b,c,ab,ac,bc,abc
思路：分两步：n-1个字符串中长度为m-1的组合；n-1个字符的长度中长度为m的组合
eg:3个字符abc，组成长度为1的组合 a -> (b,c)中取长度0的组合；(b,c)中取长度1的组合
(b,c)中取长度1的组合 b -> (c)中取长度0的组合; (c)中取长度1的组合
最终就是a,b,c
递归出口：长度=0或者字符串到达末尾' '。
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#include<iostream>
#include<string>
using namespace std;
void Combination(char* str,int length);
void combination(char* str,int length,string result);
void test(char* str,int length);
void test(char* str,int length)
{
if(NULL==str)
cout<<"origin str is NULL"<<endl;
else
cout<<"origin str="<<str<<endl;
Combination(str,length);
}
void combination(char* str,int length,string result)
{
if(*str==' '||length==0)
{
cout<<result<<endl;
return ;
}
string tmp = result + *str;
if( strlen(str+1)>=length-1 )
{
combination(str+1,length-1,tmp);
}
if(strlen(str+1)>=length)
{
combination(str+1,length,result);
}
}
void Combination(char* str,int length)
{
if(NULL==str||length<=0)
return;
string res;
for(int idx = 1;idx<=length;++idx)
{
combination(str,idx,res);
}
}
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int main()
{
char str[] = "abc";
test(str,strlen(str));
cout<<endl;
char str2[] = "ab";
test(str2,strlen(str2));
cout<<endl;
char* str3 = NULL;
test(str3,0);
cout<<endl;
char str4[] = "";
test(str4,1);
cout<<endl;
char str5[] = "a";
test(str5,2);
cout<<endl;
system("pause");
}
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