20. Valid Parentheses

深藏阁楼爱情的钟 2022-05-30 10:11 118阅读 0赞

> 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.

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**题目描述**：给出一组括号，括号是不是一一对应的，比如 ( ) 或者 \{ \} \[ \] ( )算合法的括号，（\] 或者 ( \[ ) \] 都不合法。

**思路**：这道题用loop 就非常的麻烦，但是用栈就是非常的简单，其实这道题就是考察的对栈的使用  
具体的思路如下：将前括号压入栈中，如果栈顶遇到与之相匹配后括号就弹出，最后检查栈是否为空，如果为空那么就说明全部为合法的括号

public boolean isValid(String s) {
        Stack<Character> stack = new Stack<>();
        char[] arr = s.toCharArray();
        stack.push(arr[0]);
    
        for (int i = 1; i < s.length(); i++){
            if (!stack.isEmpty() && isMatch(stack.peek(),arr[i])){
                stack.pop();
            }else {
                stack.push(arr[i]);
            }
        }
    
        return stack.isEmpty();
    
    
    }
    
    public boolean isMatch(char a, char b){
        return (a == '(' && b == ')') || (a == '[' && b == ']') || (a == '{' && b == '}');
    }

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**导论区的写法更加简练**

Stack<Character> stack = new Stack<Character>();
        for (char c : s.toCharArray()) {
            if (c == '(')
                stack.push(')');
            else if (c == '{')
                stack.push('}');
            else if (c == '[')
                stack.push(']');
            else if (stack.isEmpty() || stack.pop() != c)
                return false;
        }
        return stack.isEmpty();