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.

算法：

1 准备两个指针first, second

2 先让fisrt走n步

3 让fisrt和second同时走直到first遇到结尾

4 要用一个temp指针来记录second前一个node，用来删除second用。

/**
* 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 *first = head, *second = head;
ListNode *temp = second;

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

while(first) {
first = first -> next;
temp = second;
second = second -> next;
}

if(second == head) {
head = head -> next;
}
else {
temp -> next = second -> next;
}

return head;
}
};