15 - 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.

solution: 需考虑如下情况

1： head为空或n<= 0;

2： 当n为1时，即删除尾结点（同时考虑如果只存在一个结点）;

3： 若n大于整个链表长度，在循环得到距离head为n的结点时需要注意；

4： 遍历到倒数第n个结点了，若此结点为头结点，那么需返回head->next；

5：这个时候就可以处理最普遍的情况了，把后一个结点的值覆盖此结点，再删除后一个结点，即可完成删除倒数nth结点操作。

/**
* 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 || n <= 0)
return NULL;
else if(n == 1)
{
if(head -> next == NULL)
return NULL;
ListNode* temp = head;
while(temp->next->next != NULL)
temp = temp -> next;
temp->next = NULL;
return head;
}

ListNode* nthNode = head;
int i = 1;
while(i < n)
{
if(nthNode->next != NULL)
nthNode = nthNode -> next;
else
return NULL;
i++;
}

ListNode* node = head;

while(nthNode->next != NULL)
{
node = node -> next;
nthNode = nthNode -> next;
}

if(node == head)
return head->next;

node -> val = node -> next -> val;
node -> next = node -> next -> next;

return head;
}
};