程序媛詹妮弗
终身学习

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

验证有效括号字符串

 

思路

recursion

类似Leetcode 22 Generate Parenthesis 思路

 

代码

 1 class Solution {
 2     public boolean checkValidString(String s) {
 3         return check(s, 0, 0);
 4     }
 5     
 6     private boolean check(String s, int start, int count) {
 7         if (count < 0) return false;
 8         
 9         for (int i = start; i < s.length(); i++) {
10             char c = s.charAt(i);
11             if (c == '(') {
12                 count++;
13             }
14             else if (c == ')') {
15                 if (count <= 0) return false;
16                 count--;
17             }
18             else if (c == '*') {
19                 //1. *  for '(' --> (*))  
20                 //2. *  for ')' --> ((*)  
21                 //3. *  for empty string -->  (*)   
22                 return check(s, i + 1, count + 1) || check(s, i + 1, count - 1) || check(s, i + 1, count);
23             }
24         }
25         
26         return count == 0;
27     }
28 
29 }

 

posted on 2018-10-22 07:03  程序媛詹妮弗  阅读(116)  评论(0编辑  收藏  举报