Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
具体代码如下:
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode first = new ListNode(0);
        ListNode last = first;
        
        ListNode p = head;
        
        while(head != null){
        	while(head.next != null){//p不动,head后移直到head.next与p不相等
        		if(p.val == head.next.val){
        			head = head.next;//相等循环
        		}else{
        			break;//不相等结束
        		}
        	}
        	if(p == head){//只有一个
        		last.next = p;//添加
        		last = last.next;
        	}
        	p = head = head.next;//有多个则不添加
        	last.next = null;//去掉关联
        }
		return first.next;
    }
}
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leetCode 82.Remove Duplicates from Sorted List II (删除排序链表的重复II) 解题思路和方法
原文地址:http://blog.csdn.net/xygy8860/article/details/47002063