From 55b2318fce5be9fbeb05e0b1c29878db1a10e2bf Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Fri, 5 Oct 2018 23:14:33 +0800
Subject: [PATCH] Create 20. Valid Parentheses.md

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 20. Valid Parentheses.md | 30 ++++++++++++++++++++++++++++++
 1 file changed, 30 insertions(+)
 create mode 100644 20. Valid Parentheses.md

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+# [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/description/)
+# 思路
+用一个栈来存放左括号，每次遇到左括号就将其入栈，遇到右括号就查看是否与栈顶元素配对，若能配对则pop栈顶元素，继续下一循环，否则返回false。
+退出循环后，若栈不空，说明还剩下未配对的左括号，则应该返回false。   
+时间复杂度O(n), 空间复杂度O(n)
+# C++
+```
+class Solution {
+private:
+    bool isLegal(const char& a, const char&b){
+        if(a == '(' && b == ')') return true;
+        if(a == '[' && b == ']') return true;
+        if(a == '{' && b == '}') return true;
+        return false;
+    }
+public:
+    bool isValid(string s) {
+        stack<char>stk;
+        for(int i = 0; i < s.size(); i++){
+            if(s[i] == ')' || s[i] == '}' || s[i] == ']'){
+                if(stk.empty() || !isLegal(stk.top(), s[i])) return false;
+                stk.pop();
+            }
+            else stk.push(s[i]);
+        }
+        if(!stk.empty()) return false;
+        return true;
+    }
+};
+```