剑指offer 22 链表中倒数第k个节点

  • 分析

      双指针,让一个指针先走n步,然后快指针和慢指针同时向后移动,直到快指针移动到链表尾。

  • 代码
/**
 * 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;
        while(k > 0 && fast != nullptr){
            fast = fast -> next;
            k--;
        }

        ListNode* slow = head;

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

        return slow;
    }
};

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转载自blog.csdn.net/xiaoan08133192/article/details/109004493