LeetCode Remove Duplicates from Sorted List II

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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.

思路：如果当前的节点的值=下一个节点的值的话就一直删掉，注意细节，每次我们都只是确定了一个节点，所以pre.next每次都要置null。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
    	if (head == null) return head;
    	ListNode cur = head;
    	ListNode root = new ListNode(Integer.MAX_VALUE);
    	ListNode pre = root;
    	
    	while (cur != null && cur.next != null) {
    		if (cur.val == cur.next.val) {
    			int v = cur.val;
    			while (cur.val == v) {
    				cur = cur.next;
    				if (cur == null) return root.next;
    			}
    		} else {
    			pre.next = cur;
    			pre = pre.next;
    			cur = cur.next;
    			pre.next = null;
    		}
    	}
    	
    	pre.next = cur;
    	return root.next;
    }
}

LeetCode Remove Duplicates from Sorted List II

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原文地址：http://blog.csdn.net/u011345136/article/details/44979991