Parentheses - 20. Valid Parentheses

20. Valid Parentheses

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

An input string is valid if:

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

Note that an empty string is also considered valid.

Example 1:

Input: "()" Output: true

思路：

使用栈来做，遇到左括号入栈，遇到右括号出栈，如果不匹配就不合法。

代码：

java：

class Solution {

    /*public boolean isValid(String s) {

        if (s == null || s.length() == 0) return true;
        Stack<Character> stack = new Stack<>();
        
        for (Character ch : s.toCharArray()) {
            if (ch == '(') {
                stack.push(')');
            } else if ( ch == '[') {
                stack.push(']');
            } else if ( ch == '{') {
                stack.push('}');
            } else {
                if (stack.isEmpty() || stack.pop() != ch) return false;
            }
        }
        
        return stack.isEmpty();
    }*/
    
    public boolean isValid(String s) {
        if (s == null || s.length() == 0) return true;
        
        // use char[] as stack api
        char[] stack = new char[s.length()];
        int top = 0;
        
        for (char ch : s.toCharArray()) {
            if (ch == '(' || ch == '{' || ch == '[') {
                stack[top++] = ch;
            } else if (ch == ')' && top > 0 && stack[top-1] == '(') {
                top--;
            } else if (ch == '}' && top > 0 && stack[top-1] == '{') {
                top--;
            } else if (ch == ']' && top > 0 && stack[top-1] == '[') {
                top--;
            } else {
                return false;
            }
        }
        
        return top == 0; // 栈为空
    }
}