UVA 673 (13.08.17)
| Parentheses Balance |
You are given a string consisting of parentheses () and []. Astring of this type is said to be correct:
- (a)
- if it is the empty string
- (b)
- if A and B are correct, AB is correct,
- (c)
-
if A is correct,
(A
) and
[A
] is correct.
Write a program that takes a sequence of strings of this type and checktheir correctness. Your program can assume that the maximum stringlength is 128.
Input
The file contains a positive integer
n and a sequence of
n strings ofparentheses
() and
[], one string a line.
Output
A sequence of
Yes or
No on the output file.
Sample Input
3
([])
(([()])))
([()[]()])()
Sample Output
Yes
No
Yes
题意: 匹配小括号和中括号~
做法及其注意点:
一种情况是遇到 '(' 或 '[', 那么直接放到栈里不解释~
另一种情况是遇到 ')' 或 ']', 进行判定, 是否前一个是 '(' 或 '[' (注意了, 此时栈也许是空的), 若是, 弹出栈~
最后判断是否为空~
AC代码:
#include<stdio.h>
#include<stack> using namespace std; int main() {
int T;
char ch;
scanf("%d", &T);
getchar();
while(T--) {
stack <char> sta; while(scanf("%c", &ch)) {
if(ch == '\n')
break;
if(sta.empty()) {
sta.push(ch);
continue;
}
if(ch == '(')
sta.push(ch); if(ch == ')') {
if(sta.top() == '(')
sta.pop();
else
sta.push(ch);
}
if(ch == '[')
sta.push(ch); if(ch == ']') {
if(sta.top() == '[')
sta.pop();
else
sta.push(ch);
}
} if(sta.empty())
printf("Yes\n");
else
printf("No\n");
}
return 0;
}
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