19. Remove Nth Node From End of List

 

Given a linked list, remove the n th node from the end of list and return its head.

For example,

   Given linked list: **1- >2->3->4->5**, and **_n_ = 2**.

   After removing the second node from the end, the linked list becomes **1- >2->3->5**.

Note:
Given n will always be valid.
Try to do this in one pass.

 

这道题让我们移除链表倒数第N个节点,限定n一定是有效的,即n不会大于链表中的元素总数。还有题目要求一次遍历解决问题,那么就得想些比较巧妙的方法了。比如首先要考虑的时,如何找到倒数第N个节点,由于只允许一次遍历,所以不能用一次完整的遍历来统计链表中元素的个数,而是遍历到对应位置就应该移除了。那么就需要用两个指针来帮助解题,pre 和 cur 指针。首先 cur 指针先向前走N步,如果此时 cur 指向空,说明N为链表的长度,则需要移除的为首元素,那么此时返回 head->next 即可,如果 cur 存在,再继续往下走,此时 pre 指针也跟着走,直到 cur 为最后一个元素时停止,此时 pre 指向要移除元素的前一个元素,再修改指针跳过需要移除的元素即可,参见代码如下:

 

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

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/19

 

类似题目:

Linked List Cycle

Linked List Cycle II

 

参考资料:

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/8812/My-short-C%2B%2B-solution

https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/8804/Simple-Java-solution-in-one-pass

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation