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面试题 18. 删除链表的节点

题目描述

给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。

返回删除后的链表的头节点。

注意：此题对比原题有改动

示例 1:

输入: head = [4,5,1,9], val = 5
输出: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.

示例 2:

输入: head = [4,5,1,9], val = 1
输出: [4,5,9]
解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.

 

说明：

• 题目保证链表中节点的值互不相同
• 若使用 C 或 C++ 语言，你不需要 free 或 delete 被删除的节点

解法

方法一：模拟

我们先创建一个虚拟头节点 dummy，令 dummy.next = head，然后创建一个指针 cur 指向 dummy。

遍历链表，当 cur.next.val == val 时，将 cur.next 指向 cur.next.next，然后跳出循环。

最后返回 dummy.next 即可。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为链表的长度。

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def deleteNode(self, head: ListNode, val: int) -> ListNode:
        dummy = cur = ListNode(0, head)
        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
                break
            cur = cur.next
        return dummy.next

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        ListNode dummy = new ListNode(0, head);
        for (ListNode cur = dummy; cur.next != null; cur = cur.next) {
            if (cur.next.val == val) {
                cur.next = cur.next.next;
                break;
            }
        }
        return dummy.next;
    }
}

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteNode(ListNode* head, int val) {
        ListNode* dummy = new ListNode(0, head);
        for (ListNode* cur = dummy; cur->next; cur = cur->next) {
            if (cur->next->val == val) {
                cur->next = cur->next->next;
                break;
            }
        }
        return dummy->next;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteNode(head *ListNode, val int) *ListNode {
	dummy := &ListNode{0, head}
	for cur := dummy; cur.Next != nil; cur = cur.Next {
		if cur.Next.Val == val {
			cur.Next = cur.Next.Next
			break
		}
	}
	return dummy.Next
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} val
 * @return {ListNode}
 */
var deleteNode = function (head, val) {
    const dummy = new ListNode(0, head);
    for (let cur = dummy; cur.next; cur = cur.next) {
        if (cur.next.val == val) {
            cur.next = cur.next.next;
            break;
        }
    }
    return dummy.next;
};

Rust

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn delete_node(mut head: Option<Box<ListNode>>, val: i32) -> Option<Box<ListNode>> {
        let mut cur = &mut head;
        while let Some(node) = cur {
            if node.val == val {
                *cur = node.next.take();
                break;
            }
            cur = &mut cur.as_mut().unwrap().next;
        }
        head
    }
}

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode DeleteNode(ListNode head, int val) {
        ListNode dummy = new ListNode(0, head);
        for (ListNode cur = dummy; cur.next != null; cur = cur.next) {
            if (cur.next.val == val) {
                cur.next = cur.next.next;
                break;
            }
        }
        return dummy.next;
    }
}

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