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: 

 1 class Solution {
 2     public int scoreOfParentheses(String S) {
 3         int cur = 0;
 4         Stack<Integer> stk = new Stack<>();
 5         for(char c : S.toCharArray()){
 6             if(c == '('){
 7                 stk.push(cur);
 8                 cur = 0;
 9             }else{
10                 cur = stk.pop() + Math.max(2*cur, 1);
11             }
12         }
13         
14         return cur;
15     }
16 }

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:

 1 class Solution {
 2     public int scoreOfParentheses(String S) {
 3         int res = 0;
 4         int l = 0;
 5         for(int i = 0; i<S.length(); i++){
 6             char c = S.charAt(i);
 7             if(c == '('){
 8                 l++;
 9             }else{
10                 l--;
11             }
12             
13             if(c == ')' && S.charAt(i-1) == '('){
14                 res += 1<<l;
15             }
16         }
17         
18         return res;
19     }
20 }