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headstruct 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dummyreturn head;
}