LeetCode 856. Score of Parentheses

原题链接在这里：https://leetcode.com/problems/score-of-parentheses/

题目：

Given a balanced parentheses string S, compute the score of the string based on the following rule:

• () has score 1
• AB has score A + B, where A and B are balanced parentheses strings.
• (A) has score 2 * A, where A is a balanced parentheses string.

Example 1:

Input: "()"
Output: 1

Example 2:

Input: "(())"
Output: 2

Example 3:

Input: "()()"
Output: 2

Example 4:

Input: "(()(()))"
Output: 6

Note:

1. S is a balanced parentheses string, containing only ( and ).
2. 2 <= S.length <= 50

题解：

When encountering (, push current result to stack and set cur to 0.

When encountering ), pop the stack and plus max(current value * 2, 1). Assign it back to current number.

Time Complexity: O(n). n = S.length().

Space: O(n).

AC Java: 

class Solution {
    public int scoreOfParentheses(String S) {
        int cur = 0;
        Stack<Integer> stk = new Stack<>();
        for(char c : S.toCharArray()){
            if(c == '('){
                stk.push(cur);
                cur = 0;
            }else{
                cur = stk.pop() + Math.max(2*cur, 1);
            }
        }
        
        return cur;
    }
}

When encounter (), we care how many layers of parentheses are outside of this ().

We could have l to accumlate this. (, l++. ), l--.

And when (), accumlate 1<<l into result.

Time Complexity: O(n). n = S.length().

Space: O(1).

AC Java:

class Solution {
    public int scoreOfParentheses(String S) {
        int res = 0;
        int l = 0;
        for(int i = 0; i<S.length(); i++){
            char c = S.charAt(i);
            if(c == '('){
                l++;
            }else{
                l--;
            }
            
            if(c == ')' && S.charAt(i-1) == '('){
                res += 1<<l;
            }
        }
        
        return res;
    }
}