题解 | #牛牛的递增之旅#
牛牛的递增之旅
https://www.nowcoder.com/practice/e463addab7d548819d6b6483335651b5
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* ListNode(int x) : val(x), next(nullptr) {}
* };
*/
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param head ListNode类
* @return ListNode类
*/
ListNode* removeDuplicates(ListNode* head) {
// write code here
if(nullptr == head)
{
return head;
}
ListNode* one_node = head;
ListNode* two_node = head->next;
while(nullptr != two_node)
{
if(one_node->val == two_node->val)
{
//更新后指针的位置
two_node = two_node->next;
one_node->next = two_node;
}else {
one_node = two_node;
two_node = two_node->next;
}
}
return head;
}
};