一、问题描述
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
.
二、思路
常规思路,即发现有重复的元素时,用快指针指向下一个元素,慢指针位置不变,这点非常关键!循环比较直到不等为止。
注意:在while循环条件的判断中,必须将fast指针是否为空放在前面,将fast -> val == slow -> val放在后面,否则将会发生运行时错误,这个坑踩了好久。。。
三、代码
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(head == NULL || head -> next == NULL)
return head;
ListNode *fast = head -> next;
ListNode *slow = head;
while(fast){
if(fast -> val == slow -> val){
while(fast && fast -> val == slow -> val){
slow -> next = fast ->next;
fast = slow -> next;
}
}else{
fast = fast -> next;
slow = slow -> next;
}
}
return head;
}
};