剑指 Offer 22. 链表中倒数第k个节点

原题链接：剑指 Offer 22. 链表中倒数第k个节点

 solution:
        快慢指针,快指针先走k步,然后快慢指针一起向前走，当快指针走到终点的时候，慢指针就是倒数第k个。

class Solution {
public:
    ListNode* getKthFromEnd(ListNode* head, int k) {
        ListNode *fast = head;
        ListNode *slow = head;
        while(fast != NULL && k) {
            fast = fast->next;
            k--;
        }

        while(fast != NULL) {
            fast = fast->next;
            slow = slow->next;
        }
        return slow;
    }
};