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Easy
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Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input: "()"
Output: true
Example 2:
Input: "()[]{}"
Output: true
Example 3:
Input: "(]"
Output: false
Example 4:
Input: "([)]"
Output: false
Example 5:
Input: "{[]}"
Output: true
C++：
/*
@Date : 2019-01-07 20:28:19
@Author : 酸饺子 ([email protected])
@Link : https://github.com/SourDumplings
@Version : $Id$
*/
/*
https://leetcode.com/problems/valid-parentheses/
*/
class Solution
{
public:
bool isValid(string s)
{
stack<char> S;
for (char c : s)
{
switch (c)
{
case '[':
case '{':
case '(':
S.push(c); break;
case ']':
{
if (!S.empty() && S.top() == '[')
{
S.pop();
break;
}
else
return false;
}
case '}':
{
if (!S.empty() && S.top() == '{')
{
S.pop();
break;
}
else
return false;
}
case ')':
{
if (!S.empty() && S.top() == '(')
{
S.pop();
break;
}
else
return false;
}
}
}
if (S.empty())
{
return true;
}
else
return false;
}
};
Java:
/**
* @Date : 2019-01-07 20:35:41
* @Author : 酸饺子 ([email protected])
* @Link : https://github.com/SourDumplings
* @Version : $Id$
*
* https://leetcode.com/problems/valid-parentheses/
*/
class Solution
{
public boolean isValid(String s)
{
Stack<Character> S = new Stack<Character>();
for (int i = 0; i != s.length(); ++i)
{
Character c = new Character(s.charAt(i));
if (c == '(' || c == '[' || c == '{')
{
S.push(c);
}
else if (c == ')')
{
if (!S.empty() && S.peek() == '(')
{
S.pop();
}
else
return false;
}
else if (c == ']')
{
if (!S.empty() && S.peek() == '[')
{
S.pop();
}
else
return false;
}
else if (c == '}')
{
if (!S.empty() && S.peek() == '{')
{
S.pop();
}
else
return false;
}
}
if (S.empty())
{
return true;
}
else
return false;
}
}