Meta Phone 面试心得 | 面试通关秘籍, 技术面试题库, 编程练习

Meta Phone 面试心得 | 面试通关秘籍, 技术面试题库, 编程练习

在最近的Meta面试中，面试官提出了两道编程题目，分别考察了字符串处理和树的操作。这些问题不仅需要考生对数据结构的深入理解，还需要具备在短时间内高效解决问题的能力。我们提供的面试通关秘籍和编程练习服务可以帮助你针对这些高难度问题进行专项训练，让你在面试中游刃有余。

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题目1：Balanced Parentheses

Problem Description:

Given a string with alphanumeric characters and parentheses, remove the fewest characters possible to make the parentheses balanced. You cannot modify any characters in the string.

Code:

def balance_parentheses(input_str):
    balanced = []
    open_brackets = 0
    
    for char in input_str:
        if char == '(':
            open_brackets += 1
            balanced.append(char)
        elif char == ')':
            if open_brackets > 0:
                open_brackets -= 1
                balanced.append(char)
        else:
            balanced.append(char)
    
    result = []
    for char in balanced:
        if char == '(' and open_brackets > 0:
            open_brackets -= 1
        else:
            result.append(char)
    
    return ''.join(result)

print(balance_parentheses("(a(b)c)d)"))  # Output: (a(b)c)d

题目2：Right Side View of Binary Tree

Problem Description:

Given the root of a binary tree, imagine yourself standing on the right side of it. Return the values of the nodes you can see ordered from top to bottom.

Code:

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

def right_side_view(root):
    view = []
    if not root:
        return view

    queue = [root]
    while queue:
        size = len(queue)
        for i in range(size):
            node = queue.pop(0)
            if i == size - 1:
                view.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
    
    return view

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.right = TreeNode(5)
root.right.right = TreeNode(4)

print(right_side_view(root))  # Output: [1, 3, 4]