題目描述：給一個(gè)排好序的鏈表，刪除其中的重復(fù)元素。如：

Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3

分析：簡單的快慢指針應(yīng)用，時(shí)間復(fù)雜度O(n)，空間O(1)。

代碼：

/**
 * 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) return head;
        ListNode *pre = head, *cur;
        for (cur = head -> next; cur; cur = cur -> next)
        {
            if (pre -> val == cur -> val)
            {
                pre -> next = cur -> next;
                delete cur;
            }
            else
                pre = cur;
        }
        return head;
    }
};