题目描述补充

题目：在链表中找节点

给定一个单向链表和一个特定的值 target，编写一个函数来查找链表中是否存在值为 target 的节点。如果存在，则返回该节点的指针或引用；如果不存在，则返回 null 或一个表示不存在的值（取决于使用的编程语言）。

示例

假设链表的结构定义如下（以Python为例，其他语言类似）：

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

Python 示例代码

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def searchNode(head, target):
    current = head
    while current:
        if current.val == target:
            return current
        current = current.next
    return None

# 示例使用
# 构建链表 1 -> 2 -> 3 -> 4 -> 5
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
head.next.next.next.next = ListNode(5)

# 查找值为 3 的节点
target_node = searchNode(head, 3)
if target_node:
    print(f"Found node with value {target_node.val}")
else:
    print("Node not found")

PHP 示例代码

class ListNode {
    public $val;
    public $next;

    function __construct($val = 0, $next = null) {
        $this->val = $val;
        $this->next = $next;
    }
}

function searchNode($head, $target) {
    $current = $head;
    while ($current !== null) {
        if ($current->val == $target) {
            return $current;
        }
        $current = $current->next;
    }
    return null;
}

// 示例使用
// 构建链表 1 -> 2 -> 3 -> 4 -> 5
$head = new ListNode(1);
$head->next = new ListNode(2);
$head->next->next = new ListNode(3);
$head->next->next->next = new ListNode(4);
$head->next->next->next->next = new ListNode(5);

// 查找值为 3 的节点
$targetNode = searchNode($head, 3);
if ($targetNode !== null) {
    echo "Found node with value " . $targetNode->val . "\n";
} else {
    echo "Node not found\n";
}

JavaScript 示例代码

class ListNode {
    constructor(val = 0, next = null) {
        this.val = val;
        this.next = next;
    }
}

function searchNode(head, target) {
    let current = head;
    while (current) {
        if (current.val === target) {
            return current;
        }
        current = current.next;
    }
    return null;
}

// 示例使用
// 构建链表 1 -> 2 -> 3 -> 4 -> 5
let head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);

// 查找值为 3 的节点
let targetNode = searchNode(head, 3);
if (targetNode) {
    console.log(`Found node with value ${targetNode.val}`);
} else {
    console.log("Node not found");
}

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