leetcode -- Remove Nth Node From End of List

最新推荐文章于 2020-08-05 23:20:29 发布

原创最新推荐文章于 2020-08-05 23:20:29 发布 · 468 阅读

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本文介绍了一种高效算法，用于在一次遍历中移除链表中的倒数第N个节点，并提供了两种实现方案。第一种使用两个指针的方法，第二种则采用更加简洁的指针技巧。

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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.

分析：定义两个指针pa, pb相距为n，其中pa从head开始。pa, pb从头向尾行走，当pb走到末尾时，pa即为倒数第n个数指针。

/**
 * 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) {
        ListNode *pa = head, pb = pa, pa_pre;
        int dis = 0;
        while(pb != NULL){
            if(dis++ < n)
                pb = pb->next;
            else{
                pa_pre = pa;
                pa = pa->next;
                pb = pb->next;
            }
        }
        if(pa == head)
            head = head->next;
        else
            pa_pre->next = pa->next;
        delete pa;
        return head;
    }
};

有网友更简单的程序：

//clever solution by other guy
class Solution
{
public:
    ListNode* removeNthFromEnd(ListNode* head, int n)
    {
        ListNode** t1 = &head, *t2 = head;
        for(int i = 1; i < n; ++i)
        {
            t2 = t2->next;
        }
        while(t2->next != NULL)
        {
            t1 = &((*t1)->next);
            t2 = t2->next;
        }
        *t1 = (*t1)->next;
        return head;
    }
};