博文
练习:用栈实现逆序输出26个字母(2006-05-13 12:10:00)
摘要：//实现栈类//逆序输出26个字母//2006 05 13#include <iostream.h>#include<string>class Node{public: char data; Node *next;};class Stack:public Node{ Node *Bottom; Node *Top; int SLen;public: void InitStack(); char Pop(); void Push(char e); bool StackEmpty();};void Stack::InitStack(){ Top=new Node; Bottom=Top; SLen=0;}char Stack::Pop(){ char e; Node *p; if(!(Top->next)) exit(0); else { p=Top; &nbs......
练习:用栈检测表达式括弧匹配(2006-05-13 12:02:00)
摘要：/1、实现栈类//2、检测表达式括弧匹配//2006 05 13#include <iostream.h>#include<string>class Node{public: int data; Node *next;};class Stack:public Node{ Node *Bottom; Node *Top; int SLen;public: void InitStack(); char Pop(); void Push(int e); bool StackEmpty();};void Stack::InitStack(){ Top=new Node; Bottom=Top; SLen=0;}char Stack::Pop(){ char e; Node *p; if(!(Top->next)) exit(0); else { p=Top; &nb......
练习:用栈实现10进制到8进制的转换(2006-05-13 11:59:00)
摘要：//1、实现栈类//2、实现10进制到8进制的转换//2006 05 13#include <iostream.h>#include<string>class Node{public: int data; Node *next;};class Stack:public Node{ Node *Bottom; Node *Top; int SLen;public: void InitStack(); int Pop(); void Push(int e); bool StackEmpty();};void Stack::InitStack(){ Top=new Node; Bottom=Top; SLen=0;}int Stack::Pop(){ int e; Node *p; if(!(Top->next)) exit(0); else { p=Top; &......
练习:KMP算法 字符串匹配(2006-05-08 00:16:00)
摘要：////字符串模式匹配：KMP算法////06.05.07 vc6.0调试通过//ZYQ//#include <iostream.h>#include <string>int* get_next(char *T);void main(){ char* s="hhhsauhduwdh"; char* t="hdu"; int i=0; int j=0; int S=strlen(s); int T=strlen(t); int *next=get_next(t); while(i<S&&j<T) {//判定条件i<S&&j<T说明：字符串数组下标从0开始，到字符串长度减1结束 if(j==0||s[i]==t[j]) {i++;j++;}//继续向后比较字符 else j=next[j];//模式串向后移动 } if(j>=T) &nb......
练习:字符串首尾模式匹配(2006-05-06 20:50:00)
摘要：//字符串首尾模式匹配#include <iostream.h>#include <string>void main(){ char* s="hhhsauhduwdh"; char* t="sau"; int i=0; int j=0; int Slen=strlen(s); int Tlen=strlen(t); while(i<=Slen-Tlen+1) {// if(s[i]!=t[0]){i++;} else if(s[i]==t[0]&&s[i+Tlen-1]!=t[Tlen-1]){i++;} else { int k=1; j=1; while(j<Tlen&&s[i+k]==t......
练习:简单模式字符串匹配(2006-05-06 20:47:00)
摘要：////字符串模式匹配：简单模式下的匹配//用于串的定长顺序结构表示//#include <iostream.h>#include <string>void main(){ char* s="hhhsauhduwdh"; char* t="sau"; int i=0; int j=0; int S=strlen(s); int T=strlen(t); while(i<S&&j<T) {//判定条件i<S&&j<T说明：字符串数组下标从0开始，到字符串长度减1结束 if(s[i]==t[j]){i++;j++;} else {i=i-j+2;j=1;} } if(j>=T) cout<<i-T<<endl; else cout<<"0"<<endl; //if(j>T......