Remove Nth Node From End of List 从链表中删除倒数第N的节点

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.

采用双指针思想，两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点。

/**
* 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) {
if (head == NULL)
return NULL;

ListNode *preNode = NULL;
ListNode *pNode = head;
ListNode *qNode = head;

for(int i = 0;i<n-1;i++){
qNode = qNode->next;
}

while(qNode->next){
preNode = pNode;
pNode = pNode->next;
qNode = qNode->next;
}

if(preNode == NULL){
head = pNode->next;
delete pNode;
}else{
preNode->next = pNode->next;
delete pNode;
}

return head;
}
};