做者: 李廷元
单位: 中国民用航空飞行学院
时间限制: 400 ms
内存限制: 64 MB
代码长度限制: 16 KB编程
给定一串字符,不超过100个字符,可能包括括号、数字、字母、标点符号、空格,编程检查>这一串字符中的( ) ,[ ],{ }是否匹配。函数
输入格式:
输入在一行中给出一行字符串,不超过100个字符,可能包括括号、数字、字母、标点符号、空格。code
输出格式:
若是括号配对,输出yes,不然输出no。内存
输入样例1:
sin(10+20)
字符串输出样例1:
yes
get输入样例2:
{[}]
io输出样例2:
no
#include<stdio.h> #include<stdlib.h> #define Maxsize 100 typedef struct SNode* Stack; struct SNode { char data[Maxsize]; int Top; }; Stack create (); int Push (Stack S, char x); char Pop (Stack S); int main () { char str[101]; Stack S = create (); int i = 0, flag = 1; gets(str); while(str[i] != '\0') { if(str[i] == '(' || str[i] == '[' || str[i] == '{') Push(S, str[i]); else if(str[i] == ')' || str[i] == ']' || str[i] == '}') { if(S->Top == -1) { flag = 0; printf("no\n"); break; } if(str[i] == ')') { if(S->Top == -1 || S->data[S->Top] != '(') { flag = 0; printf("no\n"); break; } else Pop(S); } if(str[i] == ']') { if(S->Top == -1 || S->data[S->Top] != '[') { flag = 0; printf("no\n"); break; } else Pop(S); } if(str[i] == '}') { if(S->Top == -1 || S->data[S->Top] != '{') { flag = 0; printf("no\n"); break; } else Pop(S); } } i++; } if(flag != 0) { if(S->Top == -1) printf("yes\n"); else printf("no\n"); } return 0; } Stack create () { Stack S = (Stack)malloc(sizeof(struct SNode)); S->Top = -1; return S; } int Push (Stack S, char x) { if(S->Top == 99) { printf("Stack Full\n"); return -1; } else { S->data[++(S->Top)] = x; return 1; } } char Pop (Stack S) { if(S->Top == -1) { printf("Stack Empty\n"); return -1; } else return (S->data[(S->Top)--]); }
一开始是使用的while((ch = getchar()) != '\n')
来读取字符,结果一直超时,我的认为多是由于调用getchar函数次数过多。di