[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.

思路：快慢指针。注意代码鲁棒性。时间复杂度O(n),空间复杂度O(1)

相关题目：《剑指offer》面试题15 

/**
 * 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 || n <= 0) return head;
    ListNode *pfast = head;
    for (int i = 1; i < n; ++i) {
        if (pfast == NULL) break; //这是为了k的值大于链表节点总数的情况，本题不存在这种情况
        pfast = pfast->next;
    }
    ListNode **pslow = &head;
    while (pfast->next != NULL) {
        pfast = pfast->next;
        pslow = &((*pslow)->next);
    }
    
    ListNode *q = *pslow;
    *pslow = q->next;
    delete q;
    
    return head;
}
};