leetcode -- 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.

分析：定义两个指针pa, pb相距为n，其中pa从head开始。pa, pb从头向尾行走，当pb走到末尾时，pa即为倒数第n个数指针。

/**
* 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) {
ListNode *pa = head, pb = pa, pa_pre;
int dis = 0;
while(pb != NULL){
if(dis++ < n)
pb = pb->next;
else{
pa_pre = pa;
pa = pa->next;
pb = pb->next;
}
}
if(pa == head)
head = head->next;
else
pa_pre->next = pa->next;
delete pa;
return head;
}
};

有网友更简单的程序：

//clever solution by other guy
class Solution
{
public:
ListNode* removeNthFromEnd(ListNode* head, int n)
{
ListNode** t1 = &head, *t2 = head;
for(int i = 1; i < n; ++i)
{
t2 = t2->next;
}
while(t2->next != NULL)
{
t1 = &((*t1)->next);
t2 = t2->next;
}
*t1 = (*t1)->next;
return head;
}
};