# [Valid Parentheses][title]

## Description

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

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

**Example 1:**

```
Input: "()"
Output: true
```

**Example 2:**

```
Input: "()[]{}"
Output: true
```

**Example 3:**

```
Input: "(]"
Output: false
```

**Example 4:**

```
Input: "([)]"
Output: false
```

**Example 5:**

```
Input: "{[]}"
Output: true
```

**Tags:** Stack, String


## 思路

题意是判断括号匹配是否正确，很明显，我们可以用栈来解决这个问题，当出现左括号的时候入栈，当遇到右括号时，判断栈顶的左括号是否何其匹配，不匹配的话直接返回 `false` 即可，最终判断是否空栈即可，这里我们可以用数组模拟栈的操作使其操作更快，有个细节注意下 `top =  1;`，从而省去了之后判空的操作和 `top - 1` 导致数组越界的错误。

```java
class Solution {
    public boolean isValid(String s) {
        char[] stack = new char[s.length() + 1];
        int top = 1;
        for (char c : s.toCharArray()) {
            if (c == '(' || c == '[' || c == '{') {
                stack[top++] = c; 
            } else if (c == ')' && stack[--top] != '(') {
                return false;
            } else if (c == ']' && stack[--top] != '[') {
                return false;
            } else if (c == '}' && stack[--top] != '{') {
                return false;
            }
        }
        return top == 1;
    }
}
```


## 结语

如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]



[title]: https://leetcode.com/problems/valid-parentheses
[ajl]: https://github.com/Blankj/awesome-java-leetcode