# 18.删除链表的节点

> 本题较为简单，熟悉链表应该达到随手就能解出的程度

## 题目

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

来源：力扣（LeetCode）

链接：<https://leetcode-cn.com/problems/shan-chu-lian-biao-de-jie-dian-lcof>

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

## 一、 思路

本题意在移除一个节点，即：

***1 -> 2 -> 3 -> 4 -> 5***

如果要移除3节点，就要将2节点的next指向4。所以在遍历链表的时候我们在找到移除目标的时候，需要直到上个节点是什么。

这里引入一个指针 `lastNode` 保存上个节点，遍历时 `currentNode` 作为指向当前节点的指针。

遇到目标节点后，将当前节点的 `next` 赋值给 `lastNode` 的 `next`，返回 `head`。

## 二、 双指针解法

![LeetCode](/files/-MflFN29INvEu1u199VO)

```swift
func deleteNode(_ head: ListNode?, _ val: Int) -> ListNode? {
    var lastNode: ListNode?
    var currentNode = head

    while let node = currentNode {
        if node.val == val {
            if let last = lastNode {
                last.next = node.next
                return head
            }
            else {
                return node.next
            }
        }

        lastNode = node
        currentNode = node.next
    }

    return head
}
```

## 三、 单指针解法

```swift
func deleteNode2(_ head: ListNode?, _ val: Int) -> ListNode? {
    if head?.val == val {
        return head?.next
    }
    var currentNode = head
    while let node = currentNode?.next {
        if node.val == val {
            currentNode?.next = node.next
            return head
        }
        else {
            currentNode = currentNode?.next
        }
    }
    return head
}
```


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