leetcode4:Permutation
#include <utility>
#include <iostream>
#include <vector>
#include <algorithm>
//the next permutation
template<class BidirIt>
bool STL_next_permutation(BidirIt first, BidirIt last)
{
if (first == last) return false;
BidirIt i = last;
if (first == --i) return false; while () {
BidirIt i1, i2; i1 = i;
--i;
if (*i < *i1) {
i2 = last;
while (!(*i < *--i2))
;
std::iter_swap(i, i2);
std::reverse(i1, last);
return true;
}
if (i == first) {
std::reverse(first, last);
return false;
}
}
}
void nextPermutation(int A[],int len)
{
STL_next_permutation(A, A+len);
} //full pemutation
void fullPerm(int A[],int m,int n)
{
if(m == n)
{
for(int i=;i<n+;i++)
std::cout << A[i] << " ";
std::cout << std::endl;
return;
}
else
{
for(int i=m;i<n+;i++)
{
std::swap(A[m], A[i]);
fullPerm(A,m+,n);
std::swap(A[m], A[i]);
}
}
} int Factorial(int n)
{
int fac=;
for(int i=;i<=n;i++)
{
fac *=i;
}
return fac;
}
//康托编码第k个序列
void CantorCode(int A[],int len,int k)
{
--k;
std::vector<std::pair<int,bool>> v;
for(int i=;i<len;i++)
{
v.emplace_back(A[i],false);
} for(int i=;i<len;i++)
{
int j;
int t=k/Factorial(len-i-);
for(j=;j<len;j++)
{
if(!v[j].second)
{
if(t==) break;
--t;
}
}
A[i]=v[j].first;
v[j].second=true;
k=k%Factorial(len-i-);
}
}
leetcode4:Permutation的更多相关文章
- Permutation Sequence
The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the p ...
- [LeetCode] Palindrome Permutation II 回文全排列之二
Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empt ...
- [LeetCode] Palindrome Permutation 回文全排列
Given a string, determine if a permutation of the string could form a palindrome. For example," ...
- [LeetCode] Permutation Sequence 序列排序
The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the p ...
- [LeetCode] Next Permutation 下一个排列
Implement next permutation, which rearranges numbers into the lexicographically next greater permuta ...
- Leetcode 60. Permutation Sequence
The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the p ...
- UVA11525 Permutation[康托展开 树状数组求第k小值]
UVA - 11525 Permutation 题意:输出1~n的所有排列,字典序大小第∑k1Si∗(K−i)!个 学了好多知识 1.康托展开 X=a[n]*(n-1)!+a[n-1]*(n-2)!+ ...
- Permutation test: p, CI, CI of P 置换检验相关统计量的计算
For research purpose, I've read a lot materials on permutation test issue. Here is a summary. Should ...
- Permutation
(M) Permutations (M) Permutations II (M) Permutation Sequence (M) Palindrome Permutation II
随机推荐
- Android Camera的使用(一) 读书笔记
原文地址 https://blog.csdn.net/junzia/article/details/52301199 拍照步骤1.添加权限2.开启相机时check一下是否有摄像头3.对预览大小.照片大 ...
- MVC v5.1 Preview 包含 web api 2.1 web pages 3.1
Includes ASP.NET MVC 5.1, Web API 2.1, and Web Pages 3.1 preview release. This was released marked a ...
- .Net Core Runtime安装说明
在开发阶段,都是直接安装.Net Core的SDK,但是在部署的时候你还是直接装SDK吗?当然直接装SDK也没什么问题,也可以少一些麻烦.但是如果你像我一样不喜欢在产线上装SDK,只想装Runtime ...
- windows 10 RelativePanel
The new RelativePanel implements a style of layout that is defined by the relationships between its ...
- 63. 不同路径 II leetcode JAVA
题目 一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为“Start” ). 机器人每次只能向下或者向右移动一步.机器人试图达到网格的右下角(在下图中标记为“Finish”). 现在 ...
- BZOJ4012 [HNOI2015]开店 (动态点分治)
Description 风见幽香有一个好朋友叫八云紫,她们经常一起看星星看月亮从诗词歌赋谈到 人生哲学.最近她们灵机一动,打算在幻想乡开一家小店来做生意赚点钱.这样的 想法当然非常好啦,但是她们也发现 ...
- LOJ#3084. 「GXOI / GZOI2019」宝牌一大堆(递推)
题面 传送门 题解 为什么又是麻将啊啊啊!而且还是我最讨厌的爆搜类\(dp\)-- 首先国士无双和七对子是可以直接搞掉的,关键是剩下的,可以看成\(1\)个雀头加\(4\)个杠子或面子 直接\(dp\ ...
- CTF中密码学一些基础【三】
本文作者:i春秋签约作家——MAX. 看看今天教程: 看着几个字符在键盘的位置,直接就是三个圈圈,圆心的三个字符就是答案 非常简单! 答案就是KEY 看题解密就好了!! 根据提示Asp encode解 ...
- Getting Started with Elastic Search in .NET
I have been working on many application during my career. Many if not all had some searching capabi ...
- JavaScript 函数声明,函数表达式,匿名函数的区别,深入理解立即执行函数(function(){…})()
function fnName(){xxxx}; // 函数声明:使用function关键字声明一个函数,在指定一个函数名. //例如:(正常,因为 提升 了函数声明,函数调用可以在函数声明之前) f ...