数据结构括号匹配-白红宇

数据结构括号匹配

阅读量：296 次

发布时间：2019-03-03

本文共 1723 字，大约阅读时间需要 5 分钟。

括号匹配，栈的应用

#include <stdio.h>#include <stdlib.h>#define true 1#define false 0#define ok 1#define error 0#define overflow -2#define STACK_INIT_SIZE 100#define STACKINCREMENT 10typedef int status;typedef struct {       char *base;    char *top;    int stacksize;}sqstack;status initstack(sqstack &s) {   //初始化stack；    s.base = (char *)malloc(STACK_INIT_SIZE * sizeof(char));    if (!s.base) exit(overflow);    s.top = s.base;    s.stacksize = STACK_INIT_SIZE;    return ok;}status push(sqstack &s, char e) {       if (s.top - s.base >= s.stacksize) {           s.base = (char *)realloc(s.base, (s.stacksize + STACKINCREMENT) * sizeof(char));        s.top = s.base + s.stacksize;        s.stacksize += STACKINCREMENT;    }    *s.top++= e;    return ok;}status pop(sqstack &s,char &e) {       if (s.base == s.top) return error;    e = *--s.top;    return ok;}status stackempty(sqstack s) {       if (s.base == s.top) return true;    else return false;}status gettop(sqstack s, char &e) {       if (s.top == s.base) return error;    e = *(s.top - 1);    return ok;}void judge(){   	sqstack S;    initstack(S);	bool flag=true;    char op, a, b;    printf("请输入你要判断的这些括号(只能是“（、）、[、]”)\n");    op = getchar();    while (op != '\n'&& flag==true) {           switch (op) {           case '[':        case '(':push(S, op); break;        case ')':gettop(S, a); if (a == '(') pop(S,b);                 else flag = false; break;        case ']':gettop(S, a); if (a == '[') pop(S,b);                 else flag = false; break;        default: break;        }        op = getchar();    }    if (stackempty(S)) flag = true;    else flag = false;    if (flag) printf("匹配正确！");    else printf("匹配错误！");}int main() {       judge();    return 0;}