leetcode: Remove Nth Node From End of List

http://oj.leetcode.com/problems/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.

思路：

两个指针p1和p2，令p2先走N - 1步，然后两个指针一起往前走，每一步保存当前的p1为prev，当p2->next为NULL时，p1就是要删除的那个节点，另prev->next = p1->next。实现的时候比较偷懒，没有考虑输入head为NULL和n为0的情况。

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode *l1 = head, *l2 = head, *prev = NULL;
        
        for (int i = 1; i < n; ++i) {
            l2 = l2->next;
        }
        
        while (true) {
            if (NULL == l2->next) {
                break;
            }
            
            prev = l1;
            l1 = l1->next;
            l2 = l2->next;
        }
        
        if (prev != NULL) {
            prev->next = l1->next;
            return head;
        }
        else {
            return l1->next;
        }
    }
};