Leetcode: Valid Parentheses

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.

分析：
遍历字符串，使用栈依次存储字符，遇到’0’，’]’，’}’出栈进行判断。
注意：’(‘和’)’只差为1，’[‘和’]’只差为2，’{‘和’}’只差为2，这可以帮助我们判读左括号和右括号是否成对。
C++示例代码：

class Solution
{
public:
    bool isValid(string s)
    {
        string::size_type length = s.length();
        //如果给定的字符个数为奇数，直接返回false
        if (length % 2)
        {
            return false;
        }
        stack<char> chstack;
        char top;//当前栈顶的元素
        //用于记录current和previous元素只差，‘(‘和‘)‘只差为1，‘[‘和‘]‘只差为2，‘{‘和‘}‘只差为2
        int distance;
        for (size_t i = 0; i < length; i++)
        {

            if (s[i] == ‘)‘ || s[i] == ‘]‘ || s[i] == ‘}‘)
            {

                if (chstack.empty())
                {
                    return false;
                }
                else
                {
                    top = chstack.top();
                    chstack.pop();
                    distance = s[i] - top;
                    //判断previous和current是否匹配
                    if (distance != 1 && distance != 2)
                    {
                        return false;
                    }
                }
            }
            else
            {
                chstack.push(s[i]);
            }
        }
        //栈为空返回true，否则肯定有不匹配的括号在栈中返回false
        return chstack.empty() ? true : false;
    }
};

C#示例代码：

public class Solution
{
    public bool IsValid(string s)
    {
        if (s.Length % 2 == 1) return false;
        Stack<char> chstack = new Stack<char>();
        char top= ‘ ‘;
        int distance = 0;
        for (int i = 0; i < s.Length; i++)
        {
            if (s[i] == ‘)‘ || s[i] == ‘]‘ || s[i] == ‘}‘)
            {
                if (chstack.Count == 0)
                {
                    return false;
                }
                else
                {
                    top= chstack.Pop();
                    distance = s[i] - top;
                    if (distance != 1 && distance != 2)
                    {
                        return false;
                    }
                }
            }
            else
            {
                chstack.Push(s[i]);
            }
        }
        return chstack.Count == 0 ? true : false;
    }
}

Python代码：
Python代码我没有使用判读左括号和右括号距离的方法

class Solution:
    # @return a boolean
    def isValid(self, s):
        length = len(s)
        stack = []
        if length % 2 != 0:
            return False
        for i in range(length):
            if s[i] == ‘)‘ or s[i] == ‘]‘ or s[i] == ‘}‘:
                if not stack:
                    return False
                else:
                    top = stack.pop()
                    if top == ‘(‘ and s[i] != ‘)‘  or top == ‘[‘ and s[i] != ‘]‘ or top == ‘}‘ and s[i] != ‘}‘:
                        return False
            else:
                stack.append(s[i])
        if stack:
            return False
        else:
            return True