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【LeetCode】Remove Nth Node From End of List (2 solutions)

Remove Nth Node From End of List

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.

 

这题的难点在于one pass

没到尾部就没法进行倒退n个节点的操作。

但是到达尾部之后,再进行倒退删除操作,就不满足one pass了

由此诞生解法一:使用数组记录所有节点的位置,通过下标计算(尾节点位置-n+1)立刻定位到需要删除的节点了

建立vector存放ListNode*,每个指针指向一个链表节点

/** * 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)     {        vector<ListNode*> v;        ListNode* cur = head;        while(cur != NULL)        {            ListNode* p = cur;            v.push_back(p);            cur = cur->next;        }        int size = v.size();        int num = size-n;        if(num == 0)//first, no pre            return head->next;        else if(num == size-1)//last but not first, no post        {            v[num-1]->next = NULL;            return head;        }        else    //pre link to post        {            v[num-1]->next = v[num+1];            return head;        }    }};

 

仔细分析之后觉得浪费空间太多。

我们只是通过尾节点位置确定需要删除节点(n-1个偏移量),不需要其他的位置信息。

只需要两个位置相差n-1的指针,当前面的指针指向尾节点时,后面的节点即指向需要删除的节点。

由此产生解法二:

/** * 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* fast = head;  //the node at the last node        ListNode* preslow = NULL;        ListNode* slow = head;  //the node to delete                int i = 1;        while(i < n)        {            fast = fast->next;            i ++;        }                while(fast->next != NULL)        {            fast = fast->next;            preslow = slow;            slow = slow->next;        }        //fast at the last node, slow at the node to delete                if(preslow == NULL)        //delete the head            return head->next;        else        {            preslow->next = slow->next;            return head;        }    }};

 

【LeetCode】Remove Nth Node From End of List (2 solutions)