LeetCode 16 Remove Nth Node From End of List

方法一：双指针容易点。

 1 public class Solution {
 2   public ListNode removeNthFromEnd(ListNode head, int n) {
 3     // Write your solution here
 4     if (head == null) {
 5       return head; 
 6     }
 7     
 8     if (n > getNum(head)) {
 9      return head; 
10     }
11     
12     ListNode dummy = new ListNode(0);
13     dummy.next = head;
14     ListNode fast = dummy, slow = dummy;
15     for (int i = 0; i <= n; i++) {
16       fast = fast.next; 
17     }
18     
19     while (fast != null) {
20       fast = fast.next;
21       slow = slow.next;
22     }
23     slow.next = slow.next.next;
24     return dummy.next;
25   }
26   
27   private int getNum(ListNode head) {
28     int count = 0;
29     while (head != null) {
30       count++;
31       head = head.next; 
32     }
33     return count;
34   }
35 }

方法二：数个数

 1 public ListNode removeNthNode(ListNode head, int n) {
 2 // Should using dummy node here because n may remove the head node.So we must keep the head.
 3 if (head == null || n == 0) {
 4     return head;
 5 }
 6 
 7 ListNode dummy = new ListNode(0);
 8 dummy.next = head;
 9 ListNode cur = dummy;
10 
11 ListNode fast = head;
12 ListNode slow = head;
13 
14 // Firstly, count the length of the linked list
15 int len = 1;
16 while(fast.next != null && fast.next.next != null) {
17     fast = fast.next.next;
18     slow = slow.next;
19     len++;
20 }
21 //          1  2  3  4  5  6  7  8 null   , for n = 5
22 //  dummy             n
23 if (fast.next == null) {
24     len = 2 * len - 1;
25 } else if (fast.next.next == null) {
26     len = 2 * len;
27 }
28     
29 // Calculate the difference of length: 8 - 5 = 3
30 int pos = len - n;
31 
32 // If we use the dummy node, then the i must start from i = 0;
33 for (int i = 0; i < pos; i++) {
34     cur = cur.next;
35 }
36 // cur points to 3, we need to delete 4
37 cur.next = cur.next.next;
38 return dummy.next;
39 }

posted on 2018-03-02 12:16 mayinmiao 阅读(88) 评论(0) 编辑收藏举报