Add solution 020

Author: yanglbme

README.md

@@ -20,6 +20,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 007 | [Reverse Integer](https://github.com/yanglbme/leetcode/tree/master/solution/007.Reverse%20Integer) | `Math` |
 | 013 | [Roman to Integer](https://github.com/yanglbme/leetcode/tree/master/solution/013.Roman%20to%20Integer) | `Math`, `String` |
 | 014 | [Longest Common Prefix](https://github.com/yanglbme/leetcode/tree/master/solution/014.Longest%20Common%20Prefix) | `String` |
+| 020 | [Valid Parentheses](https://github.com/yanglbme/leetcode/tree/master/solution/020.Valid%20Parentheses) | `String`, `Stack` |
 | 021 | [Merge Two Sorted Lists](https://github.com/yanglbme/leetcode/tree/master/solution/021.Merge%20Two%20Sorted%20Lists) | `Linked List` |
 | 053 | [Maximum Subarray](https://github.com/yanglbme/leetcode/tree/master/solution/053.Maximum%20Subarray) | `Array`, `Divide and Conquer`, `Dynamic Programming` |
 | 070 | [Climbing Stairs](https://github.com/yanglbme/leetcode/tree/master/solution/070.Climbing%20Stairs) | `Dynamic Programming` |

solution/020.Valid Parentheses/README.md

@@ -0,0 +1,83 @@
+## 有效的括号
+### 题目描述
+
+给定一个只包括 '('，')'，'{'，'}'，'['，']' 的字符串，判断字符串是否有效。
+
+有效字符串需满足：
+
+左括号必须用相同类型的右括号闭合。
+左括号必须以正确的顺序闭合。
+注意空字符串可被认为是有效字符串。
+
+示例 1:
+```
+输入: "()"
+输出: true
+```
+
+示例 2:
+```
+输入: "()[]{}"
+输出: true
+```
+
+示例 3:
+```
+输入: "(]"
+输出: false
+```
+
+示例 4:
+```
+输入: "([)]"
+输出: false
+```
+
+示例 5:
+```
+输入: "{[]}"
+输出: true
+```
+
+### 解法
+遍历 string，遇到左括号，压入栈中；遇到右括号，从栈中弹出元素，元素不存在或者元素与该右括号不匹配，返回 false。遍历结束，栈为空则返回 true，否则返回 false。
+
+```java
+class Solution {
+    public boolean isValid(String s) {
+        if (s == null || s == "") {
+            return true;
+        }
+        char[] chars = s.toCharArray();
+        int n = chars.length;
+
+        Stack<Character> stack = new Stack<>();
+        
+        for (int i = 0; i < n; ++i) {
+            char a = chars[i];
+            if (isLeft(a)) {
+                stack.push(a);
+            } else {
+                if (stack.isEmpty() || !isMatch(stack.pop(), a)) {
+                    return false;
+                }
+            }
+        }
+        
+        return stack.isEmpty();
+                
+    }
+    
+    private boolean isMatch(char a, char b) {
+        return (a == '(' && b == ')') || (a == '[' && b == ']') || (a == '{' && b == '}');
+    }
+    
+    private boolean isLeft(char a) {
+        return a == '(' || a == '[' || a == '{';
+    }
+    
+    private boolean isRight(char a) {
+        return a == ')' || a == ']' || a == '}';
+    }
+}
+```

solution/020.Valid Parentheses/Solution.java

@@ -0,0 +1,37 @@
+class Solution {
+    public boolean isValid(String s) {
+        if (s == null || s == "") {
+            return true;
+        }
+        char[] chars = s.toCharArray();
+        int n = chars.length;
+
+        Stack<Character> stack = new Stack<>();
+        
+        for (int i = 0; i < n; ++i) {
+            char a = chars[i];
+            if (isLeft(a)) {
+                stack.push(a);
+            } else {
+                if (stack.isEmpty() || !isMatch(stack.pop(), a)) {
+                    return false;
+                }
+            }
+        }
+        
+        return stack.isEmpty();
+                
+    }
+    
+    private boolean isMatch(char a, char b) {
+        return (a == '(' && b == ')') || (a == '[' && b == ']') || (a == '{' && b == '}');
+    }
+    
+    private boolean isLeft(char a) {
+        return a == '(' || a == '[' || a == '{';
+    }
+    
+    private boolean isRight(char a) {
+        return a == ')' || a == ']' || a == '}';
+    }
+}