LeetCode.678 Valid Parenthesis String

題目

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:

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].

解答

class Solution {
    public boolean checkValidString(String s) {
        // 思路：左右括號匹配，*可以作爲任何單獨的：左括號、右括號、空符號,採用兩個棧來記錄左括號、星的位置
        if(s==null||s.length()==0){
            return true;
        }
        // 定義2個棧來存儲對應下標
        Stack<Integer>leftStack=new Stack<>();
        Stack<Integer> starStack=new Stack<>();
        
        char[] chs=s.toCharArray();
        for(int i=0;i<chs.length;i++){
           char ch=chs[i];
            if(ch=='('){
                leftStack.push(i);
            }else if(ch=='*'){
                starStack.push(i);
            }else{
                // 判斷左括號是否在星括號的右側
                if(leftStack.isEmpty()&&starStack.isEmpty()){
                    return false;
                }else if(!leftStack.isEmpty()){
                    leftStack.pop();
                }else if(!starStack.isEmpty()){
                    starStack.pop();
                }
            }
        }
        
        // 判斷leftStack是否爲空，若不爲空則校驗兩個棧的棧頂位置是否存在左括號在*括號的右邊的情況，若存在則返回false
        while(!leftStack.isEmpty()){
            if(!starStack.isEmpty()){
                if(leftStack.peek()>starStack.peek()){
                    return false;
                }else{
                    leftStack.pop();
                    starStack.pop();
                }
            }else{
                return false;
            }
        }
        return true;
    }
}