lintcode: Remove Nth Node From End of List

最新推荐文章于 2024-09-05 06:26:26 发布

martin_liang

最新推荐文章于 2024-09-05 06:26:26 发布

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本文介绍了一种高效算法，用于从链表中移除倒数第N个节点。通过双指针技巧，确保了仅需遍历一次链表即可完成操作，达到了O(n)的时间复杂度要求。

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Remove Nth Node From End of List

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

Have you met this question in a real interview?

Example

Given linked list: 1->2->3->4->5->null, and n = 2.

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

Note

The minimum number of nodes in list is n.

Challenge

O(n) time

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: The head of linked list.
     */
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // write your code here
        
        if (head == NULL)
            return NULL;
            
        if (n <= 0)
            return head;
            
        ListNode fakeHead(-1);
        fakeHead.next = head;
        
        ListNode* preNode = &fakeHead;
        ListNode* curNode = head;
        ListNode* frontNode = head;
        
        
        for (int i=1; i<n; i++)
        {
            if (frontNode)
                frontNode = frontNode->next;
            else
                return head;
        }
        
        while (frontNode->next)
        {
            frontNode = frontNode->next;
            curNode = curNode->next;
            preNode = preNode->next;
        }
        
        preNode->next = preNode->next->next;
        
        return fakeHead.next;
    }
};