【删除链表的倒数第N个结点】

Problem: 19. 删除链表的倒数第 N 个结点

思路

• 假设一共有M个结点
• 移动到删除的结点需要M-N步
• 结合链表特性，为了方便删除结点，应该移动M-N-1步
• 采用双指针，fast先移动N步，slow和fast再一起移动，直到fast == NULL
• 采用虚拟头结点

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode* fast = dummyHead;
        ListNode* slow = dummyHead;
        while (n-- && fast != NULL)
        {
            fast = fast->next;
        }
        fast = fast->next; // fast多移动一步，那slow就少移动一步，方便删除元素
        while (fast != NULL) // fast和slow一块移动
        {
            slow = slow->next;
            fast = fast->next;
        }
        ListNode* tmp = slow->next;
        slow->next = slow->next->next; // 去掉第N个结点
        delete tmp;
        head = dummyHead->next;
        delete dummyHead;
        return head;
    }
};