Offer surface prove safety questions 22 (java version): the list penultimate node K

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Offer surface prove safety questions 22 (java version): the list penultimate node K

Title Description

Input a linked list, the linked list output reciprocal k-th node.

notes

• To robustness determination: the input node is null; k <= 0 are exceptions

the complexity

• Time complexity: O (n)
• Space complexity: O (1)

public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        // input check
        if(head == null || k<=0)
            return null;
        //execute 
        ListNode fast = head;
        ListNode slow = head;
        // 下面这个循环让fast位于第K个Node上; 非正常退出循环说明链表长度不足K
        for(int i = 0; i < k - 1; i++){ // 包括slow在内, 从slow到fast是K个节点; 所以fast得先向前移动K-1步
            if(fast.next == null)
                return null; // 链表长度不足K
            fast = fast.next;
        }
        while(fast.next != null){
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}

Extremely clean code

To grasp that method above, feeling largely reflects careful thinking of when general interview

public ListNode FindKthToTail(ListNode head,int k) { 
        ListNode p, q;
        p = q = head;
        int i = 0;
        for (; p != null; i++) {
            if (i >= k)
                q = q.next;
            p = p.next;
        }
        return i < k ? null : q;
    }