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Given a linked list, remove the nth node from the end of list and return its head.
The minimum number of nodes in list is n.
Given linked list: 1->2->3->4->5->null, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5->null.
Can you do it without getting the length of the linked list?
分析：
为何删除倒数第n个node,我们需要找到那个node的parent，然后parent.next = parent.next.next. 这里我们创建一个dummy node用来处理删除的是第一个node的情况。
1 /** 2 * Definition for ListNode. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int val) { 7 * this.val = val; 8 * this.next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 /** 14 * @param head: The first node of linked list. 15 * @param n: An integer. 16 * @return: The head of linked list. 17 */ 18 ListNode removeNthFromEnd(ListNode head, int n) { 19 ListNode dummyNode = new ListNode(0); 20 dummyNode.next = head; 21 22 int size = size(dummyNode); 23 if (n + 1 > size) return dummyNode.next; 24 // get parent node 25 ListNode parent = dummyNode; 26 int kth = size - n; 27 28 int count = 1; 29 while (count < kth) { 30 parent = parent.next; 31 count++; 32 } 33 34 parent.next = parent.next.next; 35 return dummyNode.next; 36 } 37 38 public int size(ListNode head) { 39 int size = 0; 40 while (head != null) { 41 size++; 42 head = head.next; 43 } 44 return size; 45 } 46 }
如果我们不允许得到list 的长度，我们可以用下面的方法。
1 public class Solution { 2 public ListNode removeNthFromEnd(ListNode head, int n) { 3 if (n <= 0) return null; 4 5 ListNode dummy = new ListNode(0); 6 dummy.next = head; 7 8 ListNode parent = dummy; 9 for (int i = 0; i < n; i++) { 10 if (head == null) { 11 return null; 12 } 13 head = head.next; 14 } 15 while (head != null) { 16 head = head.next; 17 parent = parent.next; 18 } 19 parent.next = parent.next.next; 20 return dummy.next; 21 } 22 }
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Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5636454.html
