剑指-22

https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/

• 使用最简单的思路,遍历整个链表

• 使用双指针,可以不用遍历整个链表

1.先将两个指针指向头结点
2.fast 向前移动k步,然后fast与slow就相差k步
3.return slow

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* getKthFromEnd(ListNode* head, int k) {
        ListNode *fast = head;
        ListNode *slow = head;

        while(fast && k>0){
            fast = fast->next;
            k--;
        }

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