Leetcode 20有效的括号（附JAVA Deque 和LinkedList用法）

Posted 2022-06-23 Bug挖掘机

方法一：利用栈

思路：（1）创建一个栈，在栈里依次加入字符串的括号，如果遇到的是右括号，那么将左括号入栈，如果是右括号，则将出栈一个元素，判断出栈元素，是不是刚好是当前即将入栈的匹配右元素

python版本

class Solution(object):
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """

        # The stack to keep track of opening brackets.
        stack = []

        # Hash map for keeping track of mappings. This keeps the code very clean.
        # Also makes adding more types of parenthesis easier
        mapping = ")": "(", "": "", "]": "["

        # For every bracket in the expression.
        for char in s:

            # If the character is an closing bracket
            if char in mapping:

                # Pop the topmost element from the stack, if it is non empty
                # Otherwise assign a dummy value of # to the top_element variable
                top_element = stack.pop() if stack else #

                # The mapping for the opening bracket in our hash and the top
                # element of the stack dont match, return False
                if mapping[char] != top_element:
                    return False
            else:
                # We have an opening bracket, simply push it onto the stack.
                stack.append(char)

        # In the end, if the stack is empty, then we have a valid expression.
        # The stack wont be empty for cases like ((()
        return not stack

Java版本

class Solution 
    public boolean isValid(String s) 
        int n = s.length();
        if (n % 2 == 1) 
            return false;
        

        Map<Character, Character> pairs = new HashMap<Character, Character>() 
            put(), ();
            put(], [);
            put(, );
        ;
        Deque<Character> stack = new LinkedList<Character>();
        for (int i = 0; i < n; i++) 
            char ch = s.charAt(i);
            if (pairs.containsKey(ch)) 
                if (stack.isEmpty() || stack.peek() != pairs.get(ch)) 
                    return false;
                
                stack.pop();
             else 
                stack.push(ch);
            
        
        return stack.isEmpty();
    

如果括号只有​​()​​样式，那么判断有效的括号可以用下面的方式

def vaild(input_str):
    tmp = []
    for i in input_str:

        if i == ")" and tmp[-1] == "(":
            print(tmp[-1])
            tmp.pop()
        else:
            tmp.append(i)

    if tmp == []:
        return True
    else:
        return False

print(vaild("(()())"))

扩展，关于Java的双端队列Deque和LinkedList相关知识
一个是peek和pop的用法
​​​peek()​​​ 返回队头元素，但不删除该元素。
​​​pop()​​ 返回队头元素，并删除该元素。

示例代码如下

import java.util.LinkedList;
class LinkedListPopDemo

  public static void main(String[] args) 

     // Create a LinkedList of Strings
     LinkedList<String> list = new LinkedList<String>();

     // Add few Elements
     list.add("Jack");
     list.add("Robert");
     list.add("Chaitanya");
     list.add("kate");

     // Display LinkList elements
     System.out.println("LinkedList before: "+list);

     // pop Element from list and display it
     System.out.println("Element removed: "+list.pop());

     // Display after pop operation
     System.out.println("LinkedList after: "+list);
  

LinkedList before: [Jack, Robert, Chaitanya, kate]
Element removed: Jack
LinkedList after: [Robert, Chaitanya, kate]

总结一下：
当我们只需要取出栈顶的元素进行处理（或者说我们需要先对栈顶的数据进行处理例如比较）然后根据处理的结果进行决定是否要pop()，这种情况下，我们可以先使用peek()方法，取出栈顶的值。
补充总结一下栈中的其他常用的方法：
empty( )——如果堆栈是空的，则返回true，当堆栈包含元素时，返回false；

一个是add和push的用法区别
Java LinkedList add 是加在list尾部，LinkedList push 施加在list头部，等同于addFirst

示例代码1

import java.util.LinkedList;
class LinkedListExample 

  public static void main(String[] args) 

     // Create a LinkedList of Strings
     LinkedList<String> list = new LinkedList<String>();

     // Add few Elements
     list.add("Jack");
     list.add("Robert");
     list.add("Chaitanya");
     list.add("kate");

     // Display LinkList elements
     System.out.println("LinkedList contains: "+list);

     // push Element the list
     list.push("NEW ELEMENT");

     // Display after push operation
     System.out.println(以上是关于Leetcode 20有效的括号（附JAVA Deque 和LinkedList用法）的主要内容，如果未能解决你的问题，请参考以下文章
 
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