Valid Parentheses

Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

SOLUTION1:

使用stack来解决的简单题目。所有的字符依次入栈

1. 遇到成对的括号弹栈，弹栈不成对返回false.

2. 栈为空只能压入左括号

3. 扫描完成时，栈应该为空，否则返回FALSE.

SOLUTION 2:

使用堆栈解决，比较简单。push右括号时，检查左括号在不在，如果不在，返回false,否则弹出一个左括号。

最后看栈是否为空，不为空代表括号数不对称，也要返回false;

 1 public class Solution {
 2     public boolean isValid(String s) {
 3         if (s == null) {
 4             return false;
 5         }
 6
 7         int len = s.length();
 8
 9         // bug 1: don‘t use s as the name of the stack.
10         Stack<Character> stk = new Stack<Character>();
11
12         for (int i = 0; i < len; i++) {
13             char c = s.charAt(i);
14             switch(c) {
15                 case ‘(‘:
16                 case ‘[‘:
17                 case ‘{‘:
18                     stk.push(c);
19                     break;
20                 case ‘)‘:
21                     if (!stk.isEmpty() && stk.peek() == ‘(‘) {
22                         stk.pop();
23                     } else {
24                         return false;
25                     }
26                     break;
27                 case ‘}‘:
28                     if (!stk.isEmpty() && stk.peek() == ‘{‘) {
29                         stk.pop();
30                     } else {
31                         return false;
32                     }
33                     break;
34                 case ‘]‘:
35                     if (!stk.isEmpty() && stk.peek() == ‘[‘) {
36                         stk.pop();
37                     } else {
38                         return false;
39                     }
40                     break;
41             }
42         }
43
44         return stk.isEmpty();
45     }
46 }

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/stack/IsValid.java