题解 | #删除链表的节点#
删除链表的节点
https://www.nowcoder.com/practice/f9f78ca89ad643c99701a7142bd59f5d
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
/**
* 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
*/
struct ListNode* deleteNode(struct ListNode* head, int val ) {
// write code here
if(head == NULL) {
return NULL;
}
struct ListNode* pre = (struct ListNode*)malloc(sizeof(struct ListNode)); //慢指针
pre->val = -1;
pre->next = head;
struct ListNode* str = head; //快指针
struct ListNode* phead = pre; //头结点定位
while(str) {
if(val == str->val) {
pre->next = str->next;
str->next = NULL;
return phead->next;
}
str = str->next;
pre = pre->next;
}
return NULL;
}
