Swift - 删除链表中倒数第n个节点 例：1->2->3->4->5,n = 2 ,返回 1->2->3->5

demo下载地址

问题描述

删除链表中倒数第n个节点 例：1->2->3->4->5,n = 2 ,返回 1->2->3->5

解题思路

是快行指针，这次两个指针移动的速度相同。但是一开始，第一个指针（在指向头节点之前）就落后第二个指针n节点。接着两者同时移动，当第二个指针移动到尾结点时，第一个节点的下一个节点就是我们要删除的节点

 func p_removeNthFromEnd(head: ListNode?, n: Int) -> ListNode? {
        guard let head = head else {
            return nil
        }
        
        let dummy = ListNode.init(val: 0)
        dummy.next = head
        var prev: ListNode? = dummy
        var post: ListNode? = dummy
        
        //设置后一个节点初始位置
        
        for _ in 0..<n{
            if post == nil{
                break
            }
            
            post = post!.next
        }
        
        //同时移动前后节点
        
        while post != nil && post!.next != nil {
            prev = prev!.next
            post = post!.next
        }
        
        //删除节点
        
        prev!.next = prev!.next!.next
        
        return dummy.next
    }
    

测试

 @objc func removeNthFromEnd()
    {
        let one = ListNode.init(val: 1)
        let two = ListNode.init(val: 2)
        let three = ListNode.init(val: 3)
        let four = ListNode.init(val: 4)
        let five = ListNode.init(val: 5)
        
        one.next = two
        two.next = three
        three.next = four
        four.next = five
        
        let result = self.p_removeNthFromEnd(head: one, n: 2)
        
        var dummy = result
        while dummy != nil {
            print("node val: \(dummy!.val)")
            dummy = dummy?.next
        }
    }
    

打印结果