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LeetCode19. 删除链表的倒数第 N 个结点

19. 删除链表的倒数第 N 个结点

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。

示例 1:

img

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

代码

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int size = 0;
        ListNode* dummyNode = new ListNode(-1);
        dummyNode->next = head;
        ListNode* cur = head;
        while (cur != NULL) {
            cur = cur->next;
            size++;
        }
        cur = dummyNode;
        for (int i = 0; i < size - n; i++) {
            cur = cur->next;
        }
        cur->next = cur->next->next;
        return dummyNode->next;
    }
};

image-20240129222105158


原文地址:https://blog.csdn.net/henghuizan2771/article/details/135922973

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