LeetCode Valid Parenthesis String

原题链接在这里：https://leetcode.com/problems/valid-parenthesis-string/description/

题目：

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 

Note:

1. The string size will be in the range [1, 100].

题解：

Accumlate count of parenthesis and star.

When encountering *, increment starCount.

When encountering (, first make sure there is no more starCount than parCount.

If yes, these many more * could affect final decision. like ***(((. Finally, starCount >= parCount, but should return false.

Make starCount = parCount. Then increment parCount.

When encountering ), before decrementing parCount, if both parCount and starCount are 0, that means too many ), return false.

Finally, check if starCount >= parCount. These startCount happens after (, and they could be treated as ).

Time Complexity: O(s.length()).

Space: O(1).

AC Java:

 class Solution {
     public boolean checkValidString(String s) {
         if(s == null || s.length() == 0){
             return true;
         }

         int parCount = 0;
         int starCount = 0;
         for(int i = 0; i<s.length(); i++){
             char c = s.charAt(i);
             if(c == '('){
                 if(starCount > parCount){
                     starCount = parCount;
                 }

                 parCount++;
             }else if(c == '*'){
                 starCount++;
             }else{
                 if(parCount == 0){
                     if(starCount == 0){
                         return false;
                     }

                     starCount--;
                 }else{
                     parCount--;
                 }
             }
         }

         return starCount >= parCount;
     }
 }

When encountering ( or ), it is simple to increment or decrement.

遇到"*". 可能分别对应"(", ")"和empty string 三种情况. 所以计数可能加一, 减一或者不变. 可以记录一个范围[lo, hi].

lo is decremented by * or ).

hi is incremented by * or (.

When hi < 0, that means there are too many ), return false.

If lo > 0 at the end, it means there are too many ( left, return false.

Since lo is decremented by * and ), it could be negative. Thus lo-- should only happen when lo > 0 and maintain lo >= 0.

Note: lo can't be smaller than 0. When hi < 0, return false.

Time Complexity: O(s.length()).

Space: O(1).

AC Java:

 class Solution {
     public boolean checkValidString(String s) {
         if(s == null || s.length() == 0){
             return true;
         }

         int lo = 0;
         int hi = 0;
         for(int i = 0; i<s.length(); i++){
             if(s.charAt(i) == '('){
                 lo++;
                 hi++;
             }else if(s.charAt(i) == ')'){
                 lo--;
                 hi--;
             }else{
                 lo--;
                 hi++;
             }

             if(hi < 0){
                 return false;
             }
             lo = Math.max(lo, 0);
         }

         return lo == 0;
     }
 }