递归解决全排列

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Description

A permutation is a possible ordering of a set. For example, the permutations of a set {A,B,C} include ABC, ACB, BAC, BCA, CAB, CBA.
The number of permutations of a set of N elements is N!. The example set above has 3 elements and it has a number of 3!=6 permutations.
Given a line with N different characters, please print N! lines containing all different permutations of the N characters. These N! lines are printed in dictionary order.

Input

A line with N different characters。(1<=N<=8)

Output

N! lines containing all different permutations of the N characters. These N! lines are printed in dictionary order.

Sample Input

Copy sample input to clipboard

ABCD

Sample Output

ABCDABDCACBDACDBADBCADCBBACDBADCBCADBCDABDACBDCACABDCADBCBADCBDACDABCDBADABCDACBDBACDBCADCABDCBA

这道题要求输出字母的全排列，可以运用递归解决。

下面是代码：

#include<bits/stdc++.h>
#include<string.h>
using namespace std;
void premutation(string pre,string s)
{
if(s=="") cout<<pre<<endl;
for(int i=0;i<s.size();i++)
{
string str=s;
premutation(pre+s[i],str.erase(i,1));
}

}
int main()
{
string s,pre="";
cin>>s;
premutation(pre,s);
return 0;
}

这道题用到了erase函数，我们可以看一下它的用法：

string s;

1. s.erase(2.3) 删除第二个字符后（不包括）的三个字符，即删除s[2],s[3],s[3]。eg. 12345 ，删除3.4.5，剩下1.2.

2. s.erase(s.begin()+2,s.end()-1) 删除之间的字符，是不是可以这样理解，s.begin()==s[0],s.begin()+2==s[2].同理，在下例中，s.end()-1==s[3]。eg.12345，删除34，输出125.

3. s.erase(posion) .......(这个以后来补充。。。。。）

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