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         if(head == null) return null;
15         ListNode guard = new ListNode(Integer.MIN_VALUE);
16         guard.next = head;
17         ListNode pre = guard;
18         ListNode runner = guard;
19         int count = 0;
20         while(runner.next != null){
21             if(count < n){
22                 runner= runner.next;
23             }else{
24                 runner = runner.next;
25                 pre = pre.next;
26             }
27             count++;
28         }
29             pre.next = pre.next.next;
30         return guard.next;
31     }
32 }

 

 

 

[解题思路]
经典题。双指针，一个指针先走n步，然后两个同步走，直到第一个走到终点，第二个指针就是需要删除的节点。唯一要注意的就是头节点的处理，比如，
1->2->NULL, n =2; 这时，要删除的就是头节点。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ListNode p = head;
        int totalNum = 0;
        while(p != null){
            totalNum ++;
            p = p.next;
        }
        
        
        if(totalNum == n){
            head = head.next;
            return head;
        }
        
        
        ListNode p1 = head;
        for(int i = 0; i < (totalNum - n - 1); i ++){
           // 跳出循环的时候 p1指向的是 n 左边的那个点 
            p1 = p1.next;
        }
        ListNode p2 = p1.next;
        if(p2 != null)
            p1.next = p2.next;
        else
            p1.next = null;
        
        return head;
        
    }
}

 

ref: http://www.cnblogs.com/feiling/p/3189337.html

http://fisherlei.blogspot.com/2012/12/leetcode-remove-nth-node-from-end-of.html

posted @ 2014-02-04 16:38  Razer.Lu  阅读(224)  评论(0)  编辑  收藏  举报