LeetCode每日一题【19. 删除链表的倒数第 N 个结点】

思路：快慢指针

/**
 * 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* pre = dummyHead;
        ListNode* cur = dummyHead;
        while(n-->0 && cur)    cur = cur->next;
        cur = cur->next;
        while(cur){
            cur = cur->next;
            pre = pre->next;
        }
        pre->next = pre->next->next;
        return dummyHead->next;
    }
};
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