solve problem Score Of Parentheses

Author: P-ppc

README.md

@@ -268,6 +268,7 @@ All solutions will be accepted!
 |814|[Binary Tree Pruning](https://leetcode-cn.com/problems/binary-tree-pruning/description/)|[java/py/js](./algorithms/BinaryTreePruning)|Medium|
 |150|[Evaluate Reverse Polish Notation](https://leetcode-cn.com/problems/evaluate-reverse-polish-notation/description/)|[java/py/js](./algorithms/EvaluateReversePolishNotation)|Medium|
 |341|[Flatten Nested List Iterator](https://leetcode-cn.com/problems/flatten-nested-list-iterator/description/)|[java/py/js](./algorithms/FlattenNestedListIterator)|Medium|
+|856|[Score Of Parentheses](https://leetcode-cn.com/problems/score-of-parentheses/description/)|[java/py/js](./algorithms/ScoreOfParentheses)|Medium|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/ScoreOfParentheses/README.md

@@ -0,0 +1,2 @@
+# Score Of Parentheses
+We can solve this problem by stack

algorithms/ScoreOfParentheses/Solution.java

@@ -0,0 +1,25 @@
+class Solution {
+    public int scoreOfParentheses(String S) {
+        List<String> stack = new ArrayList<String>();
+        
+        for (int i = 0; i < S.length(); i++) {
+            String c = S.substring(i, i + 1);
+            
+            int size = stack.size();
+            if (c.equals("("))
+                stack.add(c);
+            else if (stack.get(size - 1).equals("("))
+                stack.set(size - 1, String.valueOf(1));
+            else
+                stack.set(size - 2, String.valueOf(2 * Integer.valueOf(stack.remove(size - 1))));
+            
+            int v = 0;
+            while (stack.size() > 0 && !stack.get(stack.size() - 1).equals("("))
+                v += Integer.valueOf(stack.remove(stack.size() - 1));
+            if (v > 0)
+                stack.add(String.valueOf(v));
+        }
+        
+        return stack.size() > 0 ? Integer.valueOf(stack.get(0)) : 0;
+    }
+}

algorithms/ScoreOfParentheses/solution.js

@@ -0,0 +1,26 @@
+/**
+ * @param {string} S
+ * @return {number}
+ */
+var scoreOfParentheses = function(S) {
+    let stack = []
+    
+    for (let i = 0, c = S[i]; i < S.length; i++, c = S[i]) {
+        let length = stack.length
+        
+        if (c == '(')
+            stack.push(c)
+        else if (stack[length - 1] == '(')
+            stack[length - 1] = 1
+        else
+            stack[length - 2] = 2 * stack.pop()
+        
+        let v = 0
+        while (stack.length > 0 && stack[stack.length - 1] != '(')
+            v += stack.pop()
+        if (v > 0)
+            stack.push(v)
+    }
+    
+    return stack.length > 0 ? stack[0] : 0
+};

algorithms/ScoreOfParentheses/solution.py

@@ -0,0 +1,24 @@
+class Solution(object):
+    def scoreOfParentheses(self, S):
+        """
+        :type S: str
+        :rtype: int
+        """
+        stack = []
+        
+        for c in S:
+            if c == '(':
+                stack.append(c)
+            elif stack[-1] == '(':
+                stack[-1] = 1
+            else:
+                stack[-2] = 2 * stack[-1]
+                stack.pop()
+            
+            v = 0
+            while len(stack) > 0 and stack[-1] != '(':
+                v += stack.pop()
+            if v > 0:
+                stack.append(v)
+                    
+        return 0 if len(stack) == 0 else stack[0]