LeetCode---Remove Nth Node From End of List

题目链接

题意: 给出单链表头指针head和整数n, 要求删除链表的倒数第n个结点, 这里n保证是合法的输入.

 

我的思路....其实我没大明白题目建议的one pass是什么意思, 可能就是遍历一遍链表的, 不过我还是秉着能A掉就万岁的心态...我还是首先记录了链表的长度, 然后删除第len - n + 1个结点

附上代码:

/** * 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) { if (head == NULL or (head->next == NULL and n == 1)) { return head = NULL; } int len = 0, cnt = 1; ListNode *p = head; // count the length of the list

        while (p != NULL) { len++; p = p->next; } if (n == len) { return head = head->next; } p = head; // find the (len-n)th node from the head

        while (p != NULL and cnt < len-n) { cnt++; p = p->next; } p->next = p->next->next; return head; } };

后来看了discuss, 看到老外们讨论的one pass...说真的, 我没想出来这种思路. 使用两个指针, fast和slow, fast比last多走n个结点, 当fast走到链表尾部的时候, slow就是要删除的结点的前一个结点. 可是这样还是遍历了两遍啊,( 除了最后n个结点. )

附上代码:

 1 /**  2  * Definition for singly-linked list.  3  * struct ListNode {  4  * int val;  5  * ListNode *next;  6  * ListNode(int x) : val(x), next(NULL) {}  7  * };  8  */

 9 class Solution { 10 public: 11     ListNode *removeNthFromEnd(ListNode *head, int n) { 12         if (head == NULL) { 13             return head; 14  } 15         ListNode *fast = head, *slow = head; 16         int tmp_n = n; 17         // fast pointer to the Nth node

18         while (tmp_n--) { 19             fast = fast->next; 20  } 21         if (fast == NULL) { 22             return head = head->next; 23  } 24         // move fast and slow simulataneously

25         while (fast != NULL and fast->next != NULL) { 26             fast = fast->next; 27             slow = slow->next; 28  } 29         slow->next = slow->next->next; 30         return head; 31  } 32 };