LeetCode: Remove Nth Node From End of List

原创于 2012-09-30 16:43:08 发布 · 716 阅读

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#list #null #struct #function

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本文介绍了一种优化的算法来移除单链表中倒数第N个节点，确保操作在单次遍历下完成，避免了多次遍历的开销。

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Problem:

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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;
        if (n <= 0)
            return head;
        
        ListNode* pre = head;
        ListNode* aft = head;
        while(n > 1)
        {
            pre = pre->next;
            --n;
        }
        // 删除头结点
        if (pre->next == NULL)
        {
            ListNode* del = head;
            head = head->next;
            return head;
        }
        
        pre = pre->next;
        while(pre->next != NULL)
        {
            pre = pre->next;
            aft = aft->next;
        }
        
        ListNode* del = aft->next;
        aft->next = del->next;
        return head;
    }
};