题目19. Remove Nth Node From End of List
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.
1
思路:
使用两个快慢指针,一个slow,一个fast
首先fast前进n个节点
然后slow从头(head)和fast一起每次前进一个节点,当fast链表尾部的时候,
slow指向的就是倒数第n个节点
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode fast = head;
for (int i = 0; i < n; i++) {
fast = fast.next;
}
if(fast == null){
return head.next;
}
ListNode slow = head;
while(fast.next != null){
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return head;
}
}
