leetcode 678. Valid Parenthesis String 有效括号的判断

Given a string containing only three types of characters: ‘(‘, ‘)’ and ‘*’, write a function to check whether this string is valid. We define the validity of a string by these rules:

Any left parenthesis ‘(’ must have a corresponding right parenthesis ‘)’.
Any right parenthesis ‘)’ must have a corresponding left parenthesis ‘(‘.
Left parenthesis ‘(’ must go before the corresponding right parenthesis ‘)’.
‘*’ could be treated as a single right parenthesis ‘)’ or a single left parenthesis ‘(’ or an empty string.
An empty string is also valid.
Example 1:
Input: “()”
Output: True
Example 2:
Input: “(*)”
Output: True
Example 3:
Input: “(*))”
Output: True
Note:
The string size will be in the range [1, 100].

本题题意很简单，就是多添加了一个*的处理

建议和leetcode 32. Longest Valid Parentheses 最长有效括号长度 和 leetcode 20. Valid Parentheses 一起学习

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;class Solution {public:    bool checkValidString(string s)    {        stack<int> left, star;        for (int i = 0; i < s.length(); i++)        {            if (s[i] == '*')                star.push(i);            else if (s[i] == '(')                left.push(i);            else            {                if (left.empty() == false)                    left.pop();                else if (star.empty() == false)                    star.pop();                else                    return false;            }        }        while (left.empty() == false && star.empty() == false)        {            if (star.top() < left.top())                return false;            else            {                left.pop();                star.pop();            }        }        return left.empty() == true;    }};