678. Valid Parenthesis String

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本文介绍了一种检查包含括号及星号的字符串是否有效的算法。通过定义有效性的规则，利用双指针策略进行一次遍历，判断左括号与右括号是否能正确配对，并考虑星号作为括号或空字符的可能性。

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678. Valid Parenthesis String

题目大意：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:
1. Any left parenthesis '(' must have a corresponding right parenthesis ')'.
2. Any right parenthesis ')' must have a corresponding left parenthesis '('.
3. Left parenthesis '(' must go before the corresponding right parenthesis ')'.
4. '*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string.
5. An empty string is also valid.
Example 1:
```
Input: "()"
Output: True
```
Example 2:
```
Input: "(*)"
Output: True
```
Example 3:
```
Input: "(*))"
Output: True
```
题目大意：给定一个string字符，包含‘（’，‘）’，‘’三种字符，其中‘’可以表示空字符，‘（’或者‘）’，问是否满足string字符中左右括号相等。
思路：具体见解释https://discuss.leetcode.com/topic/103936/short-java-o-n-time-o-1-space-one-pass/2

代码

package String;

import java.util.Stack;

/**
* @Author OovEver
* @Date 2017/12/8 18:57
*/
public class LeetCode678 {
  public boolean checkValidString(String s) {
      int low = 0;
      int high = 0;
     for(int i=0;i<s.length();i++) {
         if (s.charAt(i) == '(') {
             low++;
             high++;
         } else if (s.charAt(i) == ')') {
             if (low > 0) {
                 low--;
             }
             high--;
         } else {
             if (low > 0) {
                 low--;
             }
             high++;
         }
         if (high < 0) {
             return false;
         }
     }
      return low == 0;
  }
}