LeetCode OJ

题目：

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.

解题思路：

先用快慢指针的思想，找到要删除的节点和其前驱（只需遍历一遍）。然后将其删除即可。

代码：

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) return NULL;

        ListNode *pre = head, *slow = head, *quick = head;
        for (int i = 1; i <= n; i++) quick = quick->next;

        while (quick != NULL) {
            pre = slow;
            slow = slow->next;
            quick = quick->next;
        }
        if (slow != head) {
            pre->next = slow->next;
            delete slow;
            slow = NULL;
        }
        else {
            head = slow->next;
            delete slow;
        }
        return head;
    }
};