题解 | #删除链表的节点#
删除链表的节点
http://www.nowcoder.com/practice/f9f78ca89ad643c99701a7142bd59f5d
/**
- 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) { // write code here ListNode* last = head->next; ListNode* front = head; if(head->val == val) return head->next; while(last != nullptr) { if(last->val == val) { front->next = last->next; } last = last->next; front = front->next; } return head; } };
