【题目】
Given a linked list, remove the nth node from the end of list and return its head.
For 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.
Note:
Given n will always be valid.
Try to do this in one pass.
【分析】
两指针法。先让first领先n-1步,然后first和second一起移动,当first到底的时候,second即为倒数第k个节点。题目需要删除,所以我们再额外记录second的前节点。
注意其中特殊情况的考虑。
【代码】
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* removeNthFromEnd(ListNode* head, int n) { if(head == NULL || n <= 0) return head; ListNode* first = head; ListNode* second = head; ListNode* pre = NULL; for(int i = 0; i < n - 1; i++) // 中间间隔N-1的距离 first = first->next; while(first->next != NULL) { pre = second; // pre是倒数第k个节点的前一个节点,用来删除 second = second->next; // second是倒数第k个节点 first = first->next; } if(pre == NULL) // 如果倒数第k个节点就是head,那么pre会保留原来的NULL,这时head直接是倒数第k个节点second的下一个节点 head = second->next; else pre->next = second->next; return head; } };
时间: 2024-11-08 14:24:37