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; }