LeetCode——Valid Parenthese

Leetcode——Valid Parenthese

#20

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.

这一题明显是使用栈来处理这种括号匹配问题。主要的思路可以理出来，给一个字符串，有以下几种情况：
1. 对于左括号，直接放入栈中
2. 对于右括号，如果栈为空，返回false；如果栈不为空，如果栈顶为匹配的左括号，则成功，将左括号出栈，如果不是匹配的左括号，返回false。

这样就可以很容易的写出代码段：

class Solution {
public:bool isValid(string s) {stack<char> stack_p;for(int i = 0;i < s.size();i++){if(s[i] == '(' || s[i] == '{' || s[i] == '[')stack_p.push(s[i]);else{if(stack_p.empty()) return false;if(s[i] == ')' && stack_p.top() != '(') return false;if(s[i] == '}' && stack_p.top() != '{')return false;if(s[i] == ']' && stack_p.top() != '[')return false; stack_p.pop();}   }return stack_p.empty();}
};

java的代码思路基本是一样的。

Java

class Solution {public boolean isValid(String s) {if(s.length() <= 1)return false;Stack<Character> st = new Stack<Character>();for(int i = 0;i < s.length();i++){if(s.charAt(i) == '('||s.charAt(i) == '{'||s.charAt(i) == '['){st.push(s.charAt(i));else{if(st.size()==0)return false;char top = st.pop();if(s.charAt(i)==')')if(top!='(')return false;else if(s.charAt(i)=='}')if(top!='{')return false;else if(s.charAt(i)==']')if(top!='[')return false;}}return st.isEmpty();
}

Python

class Solution:def isValid(self, s):""":type s: str:rtype: bool"""parenthese = []map_parenthese = {')':'(','}':'{',']':'['}for c in s:if c in map_parenthese:if map_parenthese[c] != parenthese.pop()return false

速度上，python肯定是要慢的。

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