链表中倒数第k个节点

Description

image-20210904131800934

Solution

Using fast-slow pointers, let fast pointer firstly walk k step then slow and fast walk togerther.

Code

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/**
 * 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, *slow = head;
        for(int i = 0; i < k; i++)
            fast = fast->next;
        while(fast != nullptr){
            fast = fast->next;
            slow = slow->next;
        }
        return slow;
    }
};