678. Valid Parenthesis String

My solution for this problem is using two stacks, it's very easy to understand:

    public boolean checkValidString(String s) {
        Stack<Integer> stars = new Stack<>();
        Stack<Integer> stk = new Stack<>();

        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '*')
                stars.push(i);
            else if (s.charAt(i) == '(')
                stk.push(i);
            else {
                if (!stk.isEmpty())
                    stk.pop();
                else if (!stars.isEmpty())
                    stars.pop();
                else
                    return false;
            }
        }
        while (!stk.isEmpty()) {
            if (!stars.isEmpty()) {
                if (stk.pop() > stars.pop())
                    return false;
            } else
                return false;
        }
        return true;
    }

网友的算法，不太好理解，但是算法很好：

    public boolean checkValidString2(String s) {
        int cmin = 0, cmax = 0;
        for (int i = 0; i < s.length(); ++i) {
            char c = s.charAt(i);
            if (c == '(') {
                cmax++;
                cmin++;
            } else if (c == ')') {
                cmax--;
                cmin = Math.max(cmin - 1, 0);
            } else {
                cmax++;
                cmin = Math.max(cmin - 1, 0);
            }
            if (cmax < 0)
                return false;
        }
        return cmin == 0;
    }

标签：return,String,++,else,Valid,stars,678,stk,cmin
来源： https://www.cnblogs.com/feiflytech/p/15860160.html