链表逆序 1 #include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 using namespace std; 5 #define MAXSIZE 1000010 6 7 struct node { 8 int data; 9 int next; 10 } node[MAXSIZE]; 11 12 int List[MAXSIZE]; 13 int main() { 14 int First,n,k;…

题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/664 5-2 Reversing Linked List   (25分) Given a constant KK and a singly linked list LL, you are supposed to reverse the links of every KK elements on LL. For example, given LL being 1→2→3→4→5→6, i…

程序: 1 #include <stdio.h> 2 #include <stdlib.h> 3 4 typedef int ElementType; 5 typedef struct Node *PtrToNode; 6 struct Node { 7 ElementType Data; 8 PtrToNode Next; 9 }; 10 typedef PtrToNode List; 11 12 List Read(); /* 细节在此不表 */ 13 void Print(…

模拟栈进出 方法一: 1 #include<stdio.h> 2 #define MAXSIZE 1000 3 4 typedef struct{ 5 int data[MAXSIZE]; 6 int top; 7 }SqStack; 8 9 int InitStack(SqStack *s){ 10 s->top=-1; 11 return 0; 12 } 13 14 int Push(SqStack *s,int e){ 15 if(s->top==MAXSIZE) retur…

02-线性结构3 Reversing Linked List   (25分) 时间限制:400ms 内存限制:64MB 代码长度限制:16kB 判题程序:系统默认 作者:陈越 单位:浙江大学 https://pta.patest.cn/pta/test/3512/exam/4/question/62614 Given a constant KKK and a singly linked list LLL, you are supposed to reverse the links of ever…

02-线性结构1. Reversing Linked List (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→…

02-线性结构3 Reversing Linked List(25 point(s)) Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you m…

由于最近学的是线性结构,且因数组需开辟的空间太大.因此这里用的是纯链表实现的这个链表翻转. Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4,…

题目: 7-63 查验身份证 (15 分)  一个合法的身份证号码由17位地区.日期编号和顺序编号加1位校验码组成.校验码的计算规则如下: 首先对前17位数字加权求和,权重分配为:{7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2}:然后将计算的和对11取模得到值Z:最后按照以下关系对应Z值与校验码M的值: Z:0 1 2 3 4 5 6 7 8 9 10 M:1 0 X 9 8 7 6 5 4 3 2 现在给定一些身份证号码,请你验证校验码的有效性,并输出有问题的号码.…

2. Add Two Numbers You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may ass…