LeetCode题目:https://leetcode.com/problems/valid-parentheses/
GitHub代码:https://github.com/gatieme/LeetCode/tree/master/020-ValidParentheses
CSDN题解:http://blog.csdn.net/gatieme/article/details/51090593
Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
The brackets must close in the correct order, “()” and “()[]{}” are all valid but “(]” and “([)]” are not.
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括号匹配问题
利用一个栈来保存前括号,然后有后括号来时弹出栈顶来
注意一下情况:
* []) 左右括号数目不匹配,左括号少,右括号多,因此出栈时需判断栈是否空 if(!stack.empty())
对于第一种情况,其实如果能匹配的话,那么第一个肯定字符肯定是左括号,因此我们可以直接将第一个字符入栈,这样的话
#include
#include
#include
using namespace std;
class Solution
{
public:
bool isValid(string s)
{
stack<char> st;
bool res = true;
if(s == "")
{
return true;
}
for(int i = 0; i < s.length( ); i++)
{
// 遇见左括号就入栈
if(s[i] == '(' || s[i] == '[' || s[i] == '{')
{
st.push(s[i]);
}
else if(s[i] == ')' || s[i] == ']' || s[i] == '}')
{ // 遇见右括号就出栈
// 出栈时候, 如果栈为空, 则肯定不匹配
if(st.empty( ) == true)
{
res = false;
break;
}
// 弹出栈顶元素看是否匹配
char temp = st.top();
st.pop( );
if(s[i] == ')' && temp != '(')
{
res = false;
break;
}
else if(s[i] == ']' && temp != '[')
{
res = false;
break;
}
else if(s[i] == '}' && temp != '{')
{
res = false;
break;
}
}
}
// 所有信息操作完以后如果堆栈不为空, 说明肯定不匹配
if(st.empty() != true)
{
res = false;
}
return res;
}
};
int main()
{
Solution solu;
std::cout <"]") <<std::endl;
return 0;
}