1，next_permutation(a,a+n);（找下一个）

a代表数组头地址，b代表数组尾地址，如果下一个排列存在则返回真，否者返回假

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{int num = 1,a[6]= {1,2,3,4,5};if(next_permutation(a,a+5)){for(int i=0;i<5;i++)cout<<a[i]<<" ";cout<<endl;	}else cout<<-1<<endl;if(next_permutation(a,a+5)){for(int i=0;i<5;i++)cout<<a[i]<<" ";cout<<endl;	}else cout<<-1<<endl;return 0;
}

2，prev_permutation(a,a+n);（找上一个）

a代表数组头地址，b代表数组尾地址，如果上一个排列存在则返回真，否者返回假

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{int num = 1,a[6]= {1,2,4,3,5};if(prev_permutation(a,a+5)){for(int i=0;i<5;i++)cout<<a[i]<<" ";cout<<endl;	}else cout<<-1<<endl;if(prev_permutation(a,a+5)){for(int i=0;i<5;i++)cout<<a[i]<<" ";cout<<endl;	}else cout<<-1<<endl;return 0;
}

 3，string同样适用

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{string s={'a','b','c','d','e'};if(next_permutation(s.begin(),s.end()))cout<<s<<endl;elsecout<<-1<<endl;return 0;
}