class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
if(head == nullptr) return head;
ListNode *pre, *p;
while(head != nullptr && head->val == val){
p = head;
head = head->next;
delete p;
}
pre = head;
p = head;
if(head != nullptr) head = head->next;
while(head != nullptr){
if(head->val == val){
head = head->next;
delete pre->next;
pre->next = head;
}
else{
head = head->next;
pre = pre->next;
}
}
return p;
}
};LeetCode之Remove Linked List Elements
最新推荐文章于 2022-10-28 12:04:57 发布
本文介绍了一个用于删除链表中所有等于特定值节点的解决方案,通过迭代链表并检查每个节点的值来实现。
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