一道基础的链表相关题目，在删除时对头节点进行单独处理。
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode() : val(0), next(nullptr) {}* ListNode(int x) : val(x), next(nullptr) {}* ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* removeElements(ListNode* head, int val) {while(head != NULL && head->val == val){ListNode * t = head;head = head->next;delete t;}ListNode * cur = head;while(cur != NULL && cur->next != NULL){if(cur->next->val == val){ListNode * t = cur->next;cur->next = cur->next->next;delete t;}else{cur = cur->next;}}return head;}
};
另外一种写法是设置一个虚拟节点指向头节点，这样就无需对头节点进行单独处理，最后将head指向虚拟节点的下一个节点。
class Solution {
public:ListNode* removeElements(ListNode* head, int val) {ListNode* dummyHead = new ListNode(0); // 设置一个虚拟头结点dummyHead->next = head; // 将虚拟头结点指向head，这样方便后面做删除操作ListNode* cur = dummyHead;while (cur->next != NULL) {if(cur->next->val == val) {ListNode* tmp = cur->next;cur->next = cur->next->next;delete tmp;} else {cur = cur->next;}}head = dummyHead->next;delete dummyHead;return head;}
};
)
)
