算法学习之链表与双指针

Jackey C/C++ 579 次浏览 , , 没有评论

题目要求:

给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。

进阶:你能尝试使用一趟扫描实现吗?

 

示例 1:

输入:head = [1,2,3,4,5], n = 2
输出:[1,2,3,5]
示例 2:

输入:head = [1], n = 1
输出:[]
示例 3:

输入:head = [1,2], n = 1
输出:[1]
提示:

链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list

解题代码:

//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) {

        assert(n >= 0);

        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;

        ListNode* p = dummyHead;
        ListNode* q = dummyHead;
        for (int i = 0; i < n + 1; ++i) {
            assert(q); // 保证q不为空
            q = q->next;
        }
        while (q != NULL) {
            p = p->next;
            q = q->next;
        }

        ListNode* delNode = p->next;
        p->next = delNode->next;
        delete delNode;

        ListNode* retNode = dummyHead->next;
        delete dummyHead;

        return retNode;
    }
};

 

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