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.
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode(0);
ListNode slow = start;
ListNode fast = start;
slow.next = head; //0的下一个结点是head
for (int i = 1; i <= n + 1; i++)
fast = fast.next; //使得slow和fast之间gap为n
while (fast != null) {
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return start.next;
}
}
标签:Node,head,slow,End,19,fast,next,start,ListNode From: https://www.cnblogs.com/MarkLeeBYR/p/16853419.html