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题目描述：
给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。返回删除后的链表的头节点。
1.此题对比原题有改动
2.题目保证链表中节点的值互不相同
3.该题只会输出返回的链表和结果做对比，所以若使用 C 或 C++ 语言，你不需要 free 或 delete 被删除的节点
数据范围:
0<=链表节点值<=10000
0<=链表长度<=10000
示例：
输入：
{2,5,1,9},5
返回值：
{2,1,9}
说明：
给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 2 -> 1 -> 9
解题思路：
本题考察数据结构链表的使用。首先判断下head是不是要删除的，如果头部被删，就返回头的next；之后遍历，cur是当前结点，pre是上一个结点，如果cur的值与val一致了，就用pre连接到cur的next结点即可，相当于越过了要删除的结点，实现删除。
测试代码：
/** * 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 * @param val
int整型 * @return ListNode类 */ ListNode* deleteNode(ListNode* head, int val) {
ListNode* cur=head; ListNode* pre=head; // 判断下第一个是不是 if(head->val==val) return
head->next; cur=cur->next; // 如果找到要删除的结点，就将pre连接到cur的下一个即可，起到了删除作用 while(cur) {
if(cur->val==val) { pre->next=cur->next; return head; } cur=cur->next;
pre=pre->next; } return head; } };
