链表22--删除倒数第n个元素

墨蓝 2022-05-25 08:51 138阅读 0赞

1、题目：

Given a linked list, remove the *n*\-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.
    
    After removing the second node from the end, the linked list becomes 1->2->3->5.

2、解答：先用fast指针走n步，然后slow和fast一起走，fast走到链表结尾，slow也走到了倒数第n个结点的前一个位置；注意：当n为链表的长度时，需要特别的处理

3、C++代码

class Solution {
    public:
        ListNode* removeNthFromEnd(ListNode* head, int n) {
            if(!head)
                return head;
            
            ListNode *slow = head,*fast = head;
            
            for(int i=0;i<n;i++)
                fast = fast->next;
            
            if(!fast){
                ListNode *temp = head;
                head = head->next;
                delete temp;
                return head;
            }
            
            while(fast->next){
                slow = slow->next;
                fast = fast->next;
            }
            ListNode *temp = slow->next;
            slow->next = slow->next->next;
            delete temp;
            return head;
        }
    };

python代码

class Solution:
        def removeNthFromEnd(self, head, n):
            """
            :type head: ListNode
            :type n: int
            :rtype: ListNode
            """
            if head == None:
                return None
            
            slow,fast = head,head
            
            for i in range(0,n):
                fast = fast.next
            
            #倒数第n个元素是第一元素
            if fast == None:
                head = head.next
                return head
            
            while fast.next:
                fast = fast.next
                slow = slow.next
                
            slow.next = slow.next.next
            return head