15 - Remove Nth Node From End of List

Given a linked list, remove the n^th 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;    }};