#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <unordered_map> using namespace std; class Node {
public:
int data;
int next;
Node() : data(), next() {}
Node(int d, int n):data(d), next(n){}
}; int reverse_help(int head, int k, unordered_map<int, Node>& mem) {
if (head == -) {
return -;
} int cur = head;
int prev= -;
while (cur != -) {
if (k == ) {
break;
}
int tmp = mem[cur].next;
mem[cur].next = prev;
prev = cur;
cur = tmp; k--;
}
mem[head].next = cur;
return prev;
} int reverse(int head, int k, unordered_map<int, Node>& mem, int n) {
if (head == -) {
return -;
} int nhead = -;
int cur = head;
int pre = -; while (cur != -) {
int t = reverse_help(cur, k, mem);
if (nhead == -) {
nhead = t;
}
if (pre != -) {
mem[pre].next = t;
}
pre = cur;
cur = mem[cur].next;
n -= k;
if (n < k) {
break;
}
} return nhead;
} void print(int head, unordered_map<int, Node>& mem) {
int cur = head;
while(cur != -) {
Node& cnode = mem[cur];
if (cnode.next != -) {
printf("%05d %d %05d\n", cur, cnode.data, cnode.next);
} else {
printf("%05d %d %d\n", cur, cnode.data, cnode.next);
}
cur = mem[cur].next;
}
} int count(int head, unordered_map<int, Node>& mem) {
int cnt = ;
int cur = head;
while(cur != -) {
cnt++;
cur = mem[cur].next;
}
return cnt;
} int main() { int head, n, k;
scanf("%d%d%d", &head, &n, &k); unordered_map<int, Node> mem; for (int i=; i<n; i++) {
int addr, data, next;
scanf("%d%d%d", &addr, &data, &next);
mem.insert(make_pair(addr, Node(data, next)));
} cout<<"===="<<endl;
print(head, mem);
cout<<"===="<<endl;
head = reverse(head, k, mem, count(head, mem));
print(head, mem); return ;
}

卧槽,敢再无聊点么,输入节点数据竟然有不含在链表里的节点数据。

PAT 1074. Reversing Linked List的更多相关文章

  1. PAT 1074 Reversing Linked List[链表][一般]

    1074 Reversing Linked List (25)(25 分) Given a constant K and a singly linked list L, you are suppose ...

  2. PAT 1074. Reversing Linked List (25)

    Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elem ...

  3. PAT 甲级 1074 Reversing Linked List (25 分)(链表部分逆置,结合使用双端队列和栈,其实使用vector更简单呐)

    1074 Reversing Linked List (25 分)   Given a constant K and a singly linked list L, you are supposed ...

  4. PAT Advanced 1074 Reversing Linked List (25) [链表]

    题目 Given a constant K and a singly linked list L, you are supposed to reverse the links of every K e ...

  5. PAT (Advanced Level) 1074. Reversing Linked List (25)

    简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...

  6. PAT甲题题解-1074. Reversing Linked List (25)-求反向链表

    题意说的很清楚了,这种题的话,做的时候最好就是在纸上自己亲手模拟一下,清楚一下各个指针的情况, 这样写的时候就很清楚各个指针变量保存的是什么值. PS:一次AC哈哈,所以说自己动手在纸上画画还是很有好 ...

  7. 【PAT甲级】1074 Reversing Linked List (25 分)

    题意: 输入链表头结点的地址(五位的字符串)和两个正整数N和K(N<=100000,K<=N),接着输入N行数据,每行包括结点的地址,结点的数据和下一个结点的地址.输出每K个结点局部反转的 ...

  8. PAT甲级1074 Reversing Linked List (25分)

    [程序思路] 先根据地址按顺序读入节点,入栈,当栈里的元素个数等于k时全部出栈,并按出栈顺序保存,最后若栈不为空,则全部出栈并按出栈的稀饭顺序保存,最后输出各节点 注意:输入的节点中有可能存在无用节点 ...

  9. PAT A1074 Reversing Linked List (25 分)——链表,vector,stl里的reverse

    Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elem ...

随机推荐

  1. 关于字典的几个类--defaultdict,OrderedDict, zip()

    一.  1个键对应多个值 比如:d = {'a' : [1, 2, 3],  'b' : [4, 5]},可以使用 collections 模块中的 defaultdict 来构造这样的字典 from ...

  2. leetcode-733-Flood Fill

    题目描述: An image is represented by a 2-D array of integers, each integer representing the pixel value ...

  3. 实验一 c++简单程序设计

    一.实验内容 1.ex 2_28 (1) 用if...else判断 #include<iostream> using namespace std; int main() { char i; ...

  4. jenkins运行Python

    法一: 配置中构建执行Windows批处理命令如下 立即构建后,报错如下,提示python 不是内部或外部指令 修改Windows批处理指令如下: 再次“立即构建”则正常 法二: 安装Python插件 ...

  5. OpenFoam+CFDEM+Liggghts安装耦合

    这里安装的时间节点为:2018.10.29,安装的是目前的最新版本CFDEM,支持到与OpenFoam-5.x的耦合. 1. 先安装openfoam:https://openfoam.org/down ...

  6. Google Spanner vs Amazon Aurora: Who’ll Get the Enterprise?

    https://www.clustrix.com/bettersql/spanner-vs-aurora/ Google Spanner versus Amazon Aurora In July 20 ...

  7. 牛顿迭代法(Newton's method)

    关键词:牛顿法.牛顿迭代法.牛顿切线法.牛顿-拉弗森方法 参考:牛顿迭代法-百度百科.牛顿切线法-百度文库数学学院.牛顿切线法数值分析.非线性方程(组)的数值解法.Latex入门 https://bl ...

  8. ECharts-入门学习

    最近因项目需要要做图表,后台数据要以柱状图的形式展示,通过上网查找,感觉ECharts这个js控件挺不错的,下面就把入门知识搞一下. 一.下载ECharts控件. 地址:http://echarts. ...

  9. pull强制覆盖本地

    今天我总结的是在项目中经常用到的Git命令,上传和下拉文件. 当然在进行上传和下拉操作之前,你首先要做的就是将本地和Git库连接起来. 连接命令: git remote add origin + 你G ...

  10. linux mint 18.2 install postgresql

    https://www.postgresql.org/download/linux/ubuntu/ 1 check Xenial16.04 2  创建文件  /etc/apt/sources.list ...