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Commit 37de491

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solve problem Minimum Add To Make Parentheses Valid
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‎README.md

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|238|[Product Of Array Except Self](https://leetcode-cn.com/problems/product-of-array-except-self/description/)|[java/py/js](./algorithms/ProductOfArrayExceptSelf)|Medium|
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|896|[Monotonic Array](https://leetcode-cn.com/problems/monotonic-array/description/)|[java/py/js](./algorithms/MonotonicArray)|Easy|
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|739|[Daily Temperatures](https://leetcode-cn.com/problems/daily-temperatures/description/)|[java/py/js](./algorithms/DailyTemperatures)|Medium|
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|921|[Minimum Add To Make Parentheses Valid](https://leetcode-cn.com/problems/minimum-add-to-make-parentheses-valid/description/)|[java/py/js](./algorithms/MinimumAddToMakeParenthesesValid)|Medium|
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# Database
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|#|Title|Solution|Difficulty|
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# Minimum Add To Make Parentheses Valid
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This problem is easy to solve by stack
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class Solution {
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public int minAddToMakeValid(String S) {
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LinkedList<Character> stack = new LinkedList<Character>();
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for (char c : S.toCharArray()) {
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if (c == ')' && stack.size() > 0 && stack.peek().equals('('))
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stack.pop();
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else
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stack.push(c);
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}
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return stack.size();
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}
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}
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/**
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* @param {string} S
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* @return {number}
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*/
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var minAddToMakeValid = function(S) {
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let stack = []
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S.split('').forEach(c => {
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if (c == ')' && stack.length > 0 && stack[stack.length - 1] == '(')
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stack.pop()
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else
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stack.push(c)
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})
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return stack.length
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};
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class Solution(object):
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def minAddToMakeValid(self, S):
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"""
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:type S: str
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:rtype: int
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"""
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stack = []
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for c in S:
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if c == ')' and len(stack) > 0 and stack[-1] == '(':
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stack.pop()
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else:
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stack.append(c)
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return len(stack)

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