[LeetCode] Remove Duplicates from Sorted List II

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本文介绍如何通过迭代和比较链表节点值来删除所有重复的节点，仅保留原始链表中的唯一数值。

问题：

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.

分析：

链表的题，往往都是考察队链表的基本操作。没有高深的算法，但要一次写对也见功夫。

代码：

class Solution {
public:
	ListNode *deleteDuplicates(ListNode *head) {
		if (!head)
			return NULL;
		ListNode * dummy = new ListNode(0);
		dummy->next = head;
		ListNode * one = dummy;
		ListNode * two = head->next;
		while (two) {
			bool moved = false;
			while (two && two->val == one->next->val) {
				two = two->next;
				moved = true;
			}
			if (moved)
				one->next = two;
			else
				one = one->next;
			two = two ? two->next : NULL;
		}
		head = dummy->next;
		delete dummy;
		return head;
	}
};