LeetCode 19. Remove Nth Node From End of List

19. Remove Nth Node From End of List

Description

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.

Analysis

这道题的意思是删除链表中倒数第n个节点。

一开始的做法是先求出链表中的节点数目，然后求出倒数第n个的顺数位置。

然后遍历链表到该位置删除该节点，但是这个做法超时了。

所以我ac的方法是先用指针a找到顺数第n个节点，然后再用一个指针b从头遍历。

不难看出，当指针a走到链表尽头时，此时指针b是倒数第n个节点。删除掉即可。

Code

/**
* 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) return head;
ListNode* rear = head;
ListNode* first = head;
//rear = head;
//first = head;
for(int i = 0 ; i < n;++i){
rear = rear->next;
}

if(rear == NULL){
return head->next;
}

while(rear->next!=NULL){
first = first->next;
rear = rear->next;
}

first->next=first->next->next;
return head;
}
};