https://leetcode.com/problems/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.
/**
* 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) {
tot = 0;
if (!head) return head;
ListNode *p = head;
for (; p; p = p->next) tot++;
if (n > tot) return head;
p = head;
int cur = 1;
if (tot == n) {
head = head->next;
delete p;
} else {
while (p) {
if (cur++ == tot - n) break;
p = p->next;
}
ListNode *ret = p->next;
p->next = ret ? ret->next : NULL;
delete ret;
}
return head;
}
private:
int tot;
};
leetcode Remove Nth Node From End of List
原文：http://www.cnblogs.com/GadyPu/p/5040233.html