题目

Given a sorted linked list, delete all duplicates such that each element appear only once.

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

分析

删除链表中反复元素结点。


该题目本质非常easy。仅仅需一次遍历。须要注意的是。要释放删除的结点空间。

AC代码

/**
 * 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 *p = head, *q = p->next;
        while (p && q)
        {
            if (p->val == q->val)
            {
                ListNode *t = q;
                p->next = q->next;
                q = q->next;
                //释放要删除的结点空间
                delete t;
            }
            else{
                p = p->next;
                q = q->next;
            }//elif
        }//while
        return head;
    }
};

GitHub測试程序源代码