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

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