剑指 Offer 22. 链表中倒数第k个节点(java & python3)

java:

又是重拳出击简单题 

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        int count = 0;
        ListNode cur = head;
        while(cur != null){
            count ++;
            cur = cur.next;
        }
        for( cur = head; k < count; count--){
            cur = cur.next;
        }
        return cur;
    }
}

双指针一下子～～～：

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode fast = head;
        ListNode slow = head;

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

        while( fast != null ){
            fast = fast.next;
            slow = slow.next;
        }
        
        return slow;
    }
}

python3:

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        cur = head
        count = 0
        while cur != None:
            count += 1
            cur = cur.next
        cur = head
        while k < count :
            cur = cur.next
            count -= 1
        return cur

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        fast = head
        slow = head

        while k > 0:
            fast = fast.next
            k -= 1

        while fast != None:
            fast = fast.next
            slow = slow.next

        return slow