leetcode678- Valid Parenthesis String- medium

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].

 

1.O(n) 空间。用两个Stack<Integer>分别记录left和star出现的下标。导致失败的情况是：1.出现right的时候，left和star合起来都不够用了。2.遍历完以后left和star配对消不掉：表现为，栈顶的star下标比栈顶的left下标小 就好像*( ，或者star先被用完了（用最后返回left.isEmpty()来表现）。

2.O(1)空间。用两个int most, least记录着现在还有多少个left可用，区别是most记录把星号当left用的情况，least记录把星号当right用的情况。导致失败的情况是：1.出现right的时候，most都到0了。 2.遍历完以后，least还>0，说明最后星号不够抵消。  更新情况是这样的：1.出现left，most, least都++。2.出现right，most--, least没到0的话-- 3.出现star，most++, least没到0的话--。（23里都对least很宽容的原因是，用most到0就失败的机制，已经避免了right出现太多的情况，那这个时候出现right时least已经到0只能是因为中间出现过star帮忙消了left造成的，这时候你就不用多减了，当之前的star是空气，现在的right实打实地替代了前面那个*即可）

 

实现1：

class Solution {
    public boolean checkValidString(String s) {
        Stack<Integer> left = new Stack<>();
        Stack<Integer> star = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                left.push(i);
            } else if (c == '*') {
                star.push(i);
            } else {
                if (!left.isEmpty()) {
                    left.pop();
                } else if (!star.isEmpty()) {
                    star.pop();
                } else {
                    return false;
                }
            }
        }
        while (!left.isEmpty() && !star.isEmpty()) {
            int leftI = left.pop();
            int starI = star.pop();
            if (starI < leftI) {
                return false;
            }
        }
        return left.isEmpty();
    }
}

　　

实现2：

class Solution {
    public boolean checkValidString(String s) {
        // number of '(' when treating '*' as '('
        int most = 0;
        // number of '(' when treating '*' as ')'
        int least = 0;
        
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                least++;
                most++;
            } else if (c == ')') {
                if (least > 0) {
                    least--;
                }
                most--;
            } else {
                if (least > 0) {
                    least--;
                }
                most++;
            }
            if (most < 0) {
                return false;
            }
        }
        return least == 0;
    }
}