20190601-剑指offer22-AcWing-33. 链表中倒数第k个节点

剑指offer22-AcWing-33. 链表中倒数第k个节点

输入一个链表，输出该链表中倒数第k个结点。
注意：
k >= 0;
如果k大于链表长度，则返回 NULL;
样例
输入：链表：1->2->3->4->5 ，k=2
输出：4

思路：
  首先判断链表为空、以及链表没有k个、k<=0的情况
  然后，初始两个指针，让其中一个先走k步，第二个指针再开始走直到第一个指针为空，则第二个指针指向的位置即结果。
c++ code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* findKthToTail(ListNode* pListHead, int k) {
        if(pListHead == NULL || k == 0){
            return NULL;
        }
        ListNode* nowNode = pListHead;
        ListNode* resNode = pListHead;
        while(k > 0 && nowNode){
            nowNode = nowNode->next;
            k--;
        }
        if(k > 0){
            return NULL;
        }
        while(nowNode){
            nowNode = nowNode->next;
            resNode = resNode->next;
        }
        return resNode;
    }
};

python code:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def findKthToTail(self, pListHead, k):
        """
        :type pListHead: ListNode
        :type k: int
        :rtype: ListNode
        """
        if pListHead == None or k == 0:
            return None
            
        nowNode = pListHead
        resNode = pListHead
        while k > 0 and nowNode:
            nowNode = nowNode.next
            k -= 1
        
        if k > 0:
            return None
        
        while nowNode:
            nowNode = nowNode.next
            resNode = resNode.next
            
        return resNode