[leetcode] 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.
https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
思路1:双指针思想,两个指针相隔n-1,每次两个指针向后一步,当后面一个指针没有后继了,前面一个指针就是要删除的节点。注意:删除节点时需要删除指针的前驱pre;增加dummy head处理删除头节点的特殊情况。
public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { if (head == null) return null; ListNode dummyHead = new ListNode(-1); dummyHead.next = head; ListNode s = head; ListNode t = head; n = n - 1; for (int i = 0; i < n; i++) { t = t.next; } ListNode preS = dummyHead; while (t.next != null) { preS = preS.next; s = s.next; t = t.next; } // remove s preS.next = s.next; s.next = null; return dummyHead.next; } public static void main(String[] args) { ListNode head = new ListNode(1); head.next = new ListNode(2); head.next.next = new ListNode(3); head.next.next.next = new ListNode(4); head.next.next.next.next = new ListNode(5); ListNode newHead = new Solution().removeNthFromEnd(head, 1); } }
第二遍记录:
注意head是null的情况。
注意移动的条件是t.next!=null
public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { if(head==null) return null; ListNode dummyHead = new ListNode(-1); dummyHead.next = head; n=n-1; ListNode t=head; for(int i=0;i<n;i++) t=t.next; ListNode pre =dummyHead; while(t.next!=null){ pre=pre.next; t=t.next; } ListNode toRemove=pre.next; pre.next = toRemove.next; toRemove.next=null; return dummyHead.next; } }