19. Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

原创最新推荐文章于 2024-01-12 15:01:11 发布 · 140 阅读

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本文介绍了一种高效算法，用于删除单链表中倒数第N个节点，提供两种解决方案：一种通过两次遍历计算链表长度并删除目标节点；另一种采用双指针技巧实现单次遍历即可完成任务。

Given a linked list, remove the nth 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.

其要求为：

给定一个链表，删除倒数第n个节点，如：a->b->c->d->e->NULL, n = 2，返回：a->b->c->e

解法1：先遍历一遍计算链表长度；再遍历一遍删除倒数第n个节点，该方法需要遍历两遍链表。

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {

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

        int length = 0;
        for(ListNode* cur = tmpHead->next ; cur != NULL ; cur = cur->next )
            length ++;
        int k = length - n;
        if (k < 0)
            return NULL;
        ListNode* cur = tmpHead;
        for( int i = 0 ; i < k ; i ++ )
            cur = cur -> next;

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

        ListNode* retNode = tmpHead->next;
        delete tmpHead;
        return retNode;
    }
};

解法2：使用双指针, 对链表只遍历了一遍

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {

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

        ListNode* p = tmpHead;
        ListNode* q = tmpHead;
        for( int i = 0 ; i < n + 1 ; i ++ )
        {
            if (!q)
                return "";
            q = q->next;
        }

        while( q ){
            p = p->next;
            q = q->next;
        }

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

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

        return retNode;
    }
};