678. Valid Parenthesis String

678. Valid Parenthesis String

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].

　　看到题目后，有印象曾经也在leetcode做过括号配对的题目，但是这题比较难处理的是‘*’号，作为任意匹配字符，既可以消耗一个‘)’号，也可以作为无用字符。思考一会后，基本确定思路：

　　　　1、使用stack，入栈‘(’和‘*’号，在遇到‘)'时，出栈(’和‘*’号；

　　　　2、出栈过程中优先消耗左括号，实在没有对应的左括号时，消耗一个离栈顶最近的星号；

　　　　3、从栈顶开始找左括号的过程中，如果找到对应的做左括号，要把寻找过程中出栈的星号再次压回栈中。

　　具体代码如下，要注意栈中无元素的情况下，继续出栈，会报ArrayIndexOutOfBoundsException异常：

    public static void main(String[] args) {
        // String s = "((*)"
        String s = "(((******))";
        // String s = "(())((())()()(*)(*()(())())())()()((()())((()))(*";
        System.out.println(checkValidString(s));
    }

    public static boolean checkValidString(String s) {
        Stack<Character> chars = new Stack<>();
        int countStart = 0;
        for (int i = 0; i < s.length(); i++) {
            char curr = s.charAt(i);
            if ('(' == curr || '*' == curr) {
                chars.push(curr);
            } else if (')' == curr) {
                if (chars.size() == 0) {
                    return false;
                }
                countStart = 0;
                while (!chars.isEmpty() && chars.peek() == '*') {
                    chars.pop();
                    countStart++;
                }
                if (!chars.isEmpty()) {
                    chars.pop();
                    while (countStart-- > 0) {
                        chars.push('*');
                    }
                } else if (countStart > 0) {
                    int temp = countStart - 1;
                    while (temp-- > 0) {
                        chars.push('*');
                    }
                } else {
                    return false;
                }
            }
        }
        if (chars.isEmpty()) {
            return true;
        } else {
            countStart = 0;
            while (!chars.isEmpty()) {
                char a = chars.pop();
                if ('(' == a) {
                    if (countStart > 0) {
                        countStart--;
                    } else {
                        return false;
                    }
                } else {
                    countStart++;
                }
            }
        }
        return true;
    }