LeetCode | 19. 删除链表的倒数第 N 个结点
LeetCode | 19. 删除链表的倒数第 N 个结点
OJ链接
思路：
- 定义虚拟头节点
dummy并初始化使其指向head - 然后定义快慢指针
- 让快指针先走n步
- 然后一起走
- 最后删除倒数第n个节点
- 然后释放虚拟节点
dummy
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
//定义虚拟头节点dummy 并初始化使其指向head
struct ListNode* dummy = malloc(sizeof(struct ListNode));
dummy->val = 0;
dummy->next = head;
//定义 fast slow 双指针
struct ListNode* fast = head;
struct ListNode* slow = dummy;
for (int i = 0; i < n; ++i) {
fast = fast->next;
}
while (fast) {
fast = fast->next;
slow = slow->next;
}
slow->next = slow->next->next;//删除倒数第n个节点
head = dummy->next;
free(dummy);//删除虚拟节点dummy
return head;
}