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Remove Nth Node From End of List
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 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 public ListNode removeNthFromEnd(ListNode head, int n) {14 List<ListNode> list = new ArrayList<ListNode>();15 ListNode temp = head;16 if(null == head.next)17 return null;18 while(null != temp){19 list.add(temp);20 temp = temp.next;21 }//只要一趟，把所有节点保存起来22 23 if(list.size() - n - 1 >= 0){24 temp = list.get(list.size() - n - 1);25 temp.next = temp.next.next;26 return head;27 }28 head = head.next;29 return head;30 }31 }
ps:依然是双指针思想，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。博客园上面看到的，就没有实现了
Remove Nth Node From End of List
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