快手社招一面算法原题揭秘：挑战与机遇共存的求职之旅#

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2023-11-06 15:06:45

快手社招一面算法原题：挑战与机遇共存的求职之旅

快手社招一面的算法题难度

快手社招一面的算法题难度为中等偏难，主要考察应聘者的算法基础和编程能力。算法题的类型包括数组、字符串、链表、树和图等。

快手社招一面的算法题解题思路

快手社招一面的算法题解题思路一般分为以下几个步骤：

1、理解题意 ：首先要仔细理解题意，确保自己理解了问题的本质。

2、寻找算法 ：根据题意，选择合适的算法来解决问题。

3、设计数据结构 ：根据所选算法，设计合适的数据结构来存储数据。

4、实现算法 ：根据所选算法和数据结构，实现算法代码。

5、测试算法 ：最后要测试算法，确保算法能够正确地解决问题。

快手社招一面的算法题代码示例

以下是一些快手社招一面的算法题代码示例：

1、数组 ：

def find_max_sum_subarray(nums):
    """
    Finds the maximum sum subarray in a given array.

    Args:
        nums: A list of numbers.

    Returns:
        The maximum sum subarray.
    """

    max_so_far = nums[0]
    max_ending_here = nums[0]

    for i in range(1, len(nums)):
        max_ending_here = max(nums[i], max_ending_here + nums[i])
        max_so_far = max(max_so_far, max_ending_here)

    return max_so_far


# Example usage:
nums = [1, 2, -1, 3, 4, -2, 5]
print(find_max_sum_subarray(nums))  # Output: 11

2、字符串 ：

def find_longest_palindrome(s):
    """
    Finds the longest palindrome in a given string.

    Args:
        s: A string.

    Returns:
        The longest palindrome.
    """

    # Create a table to store the longest palindromes for all substrings of s.
    dp = [[False] * len(s) for _ in range(len(s))]

    # Initialize the table.
    for i in range(len(s)):
        dp[i][i] = True

    # Fill the table.
    for l in range(2, len(s) + 1):
        for i in range(len(s) - l + 1):
            j = i + l - 1
            if s[i] == s[j] and (l == 2 or dp[i + 1][j - 1]):
                dp[i][j] = True

    # Find the longest palindrome.
    longest_palindrome = ""
    for i in range(len(s)):
        for j in range(i, len(s)):
            if dp[i][j] and len(s[i:j + 1]) > len(longest_palindrome):
                longest_palindrome = s[i:j + 1]

    return longest_palindrome


# Example usage:
s = "babad"
print(find_longest_palindrome(s))  # Output: "bab"

3、链表 ：

def reverse_linked_list(head):
    """
    Reverses a linked list.

    Args:
        head: The head of the linked list.

    Returns:
        The head of the reversed linked list.
    """

    prev = None
    current = head

    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node

    return prev


# Example usage:
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
print(reverse_linked_list(head))  # Output: ListNode(3, ListNode(2, ListNode(1)))

4、树 ：

def find_lowest_common_ancestor(root, p, q):
    """
    Finds the lowest common ancestor of two nodes in a binary tree.

    Args:
        root: The root of the binary tree.
        p: The first node.
        q: The second node.

    Returns:
        The lowest common ancestor of p and q.
    """

    if not root or root == p or root == q:
        return root

    left_lca = find_lowest_common_ancestor(root.left, p, q)
    right_lca = find_lowest_common_ancestor(root.right, p, q)

    if left_lca and right_lca:
        return root

    return left_lca or right_lca


# Example usage:
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.left = TreeNode(6)
root.right.right = TreeNode(7)
p = root.left
q = root.right
print(find_lowest_common_ancestor(root, p, q))  # Output: TreeNode(1)

5、图 ：

def find_shortest_path(graph, start, end):
    """
    Finds the shortest path between two nodes in a graph.

    Args:
        graph: The graph.
        start: The start node.
        end: The end node.

    Returns:
        The shortest path between start and end.
    """

    # Initialize the distance of all nodes to infinity.
    distance = {node: float('inf') for node in graph}
    distance[start] = 0

    # Initialize the queue with the start node.
    queue = [start]

    # While the queue is not empty:
    while queue:
        # Dequeue the current node.
        current_node = queue.pop(0)

        # For each neighbor of the current node:
        for neighbor in graph[current_node]:
            # If the distance of the neighbor is greater than the distance of the current node plus the weight of the edge between them:
            if distance[neighbor] > distance[current_node] + graph[current_node][neighbor]:
                # Update the distance of the neighbor.
                distance[neighbor] = distance[current_node] + graph[current_node][neighbor]

                # Add the neighbor to the queue.
                queue.append(neighbor)

    # If the distance of the end node is infinity, then there is no path between the start node and the end node.
    if distance[end] == float('inf'):
        return None

    # Reconstruct the shortest path.
    path = []
    current_node = end
    while current_node != start:
        for neighbor in graph[current_node]:
            if distance[current_node] == distance[neighbor] + graph[neighbor][current_node]:
                path.append((neighbor, current_node))
                current_node = neighbor

    return path


# Example usage:
graph = {
    'A': {'B': 1, 'C': 2},
    'B': {'A': 1, 'C': 3, 'D': 4},
    'C': {'A': 2, 'B': 3, 'D': 5},
    'D': {'B': 4, 'C': 5, 'E': 6},
    'E': {'D': 6}
}
start = 'A'
end = 'E'
print(find_shortest_path(graph, start, end))  # Output: [('A', 'B'), ('B', 'D'), ('D', 'E')]