[LeetCode]20.Valid Parentheses
2017-05-15 22:15
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题目: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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解题思路是这样的:我们对字符串S中的每一个字符C,如果C不是右括号,就压入栈stack中。如果C是右括号,判断stack是不是空的,空则说明没有左括号,直接返回not valid,非空就取出栈顶的字符pre来对比,如果是匹配的,就弹出栈顶的字符,继续取S中的下一个字符;如果不匹配,说明不是valid的,直接返回。当我们遍历了一次字符串S后,注意这里还有一种情况,就是stack中还有残留的字符没有得到匹配,即此时stack不是空的,这时说明S不是valid的,因为只要valid,一定全都可以得到匹配使左括号弹出。
#include <stack>
class Solution {
public:
bool isValid(string s) {
stack<char> paren;
for (char& c : s) {
switch (c) {
case '(':
case '{':
case '[': paren.push(c); break;
case ')': if (paren.empty() || paren.top() != '(') return false; else paren.pop(); break;
case '}': if (paren.empty() || paren.top() != '{') return false; else paren.pop(); break;
case ']': if (paren.empty() || paren.top() != '[') return false; else paren.pop(); break;
default:; // pass
}
}
return paren.empty();
}
};
The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
Subscribe to see which companies asked this question.
解题思路是这样的:我们对字符串S中的每一个字符C,如果C不是右括号,就压入栈stack中。如果C是右括号,判断stack是不是空的,空则说明没有左括号,直接返回not valid,非空就取出栈顶的字符pre来对比,如果是匹配的,就弹出栈顶的字符,继续取S中的下一个字符;如果不匹配,说明不是valid的,直接返回。当我们遍历了一次字符串S后,注意这里还有一种情况,就是stack中还有残留的字符没有得到匹配,即此时stack不是空的,这时说明S不是valid的,因为只要valid,一定全都可以得到匹配使左括号弹出。
#include <stack>
class Solution {
public:
bool isValid(string s) {
stack<char> paren;
for (char& c : s) {
switch (c) {
case '(':
case '{':
case '[': paren.push(c); break;
case ')': if (paren.empty() || paren.top() != '(') return false; else paren.pop(); break;
case '}': if (paren.empty() || paren.top() != '{') return false; else paren.pop(); break;
case ']': if (paren.empty() || paren.top() != '[') return false; else paren.pop(); break;
default:; // pass
}
}
return paren.empty();
}
};
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