82. Remove Duplicates from Sorted List II

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.

这道题要考虑如果头部多个相同的时候要重新定义head，多添加一个head之前的节点，如果cur和cur->next不同的时候且pre->next不是cur就删除所有中间节点。最后返回的是new_head->next

代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == NULL || head->next == NULL) return head;
        
        ListNode *new_head = new ListNode(0);
        new_head->next = head;
        ListNode *pre = new_head;
        ListNode *cur = head;
        
        while (cur->next) {
            if (cur->next->val != cur->val) {
                if (pre->next == cur) {
                    pre = cur;
                }
                else pre->next = cur->next;//delete nodes
            }
            cur = cur->next;
        }
        
        if (pre->next != cur) pre->next = cur->next;
        return new_head->next;
    }
};