14. 链表中倒数第k个节点

题目描述

输入一个链表，输出该链表中倒数第k个结点。

题目思路:除了以下两种方法,还想到一种方法:将链表倒序,然后找到正数第k个节点,其实这种方法是不行的,因为每个节点都有两个属性:next、val,倒叙后得到的正数第k个节点,虽然val相同,但是next是不同的.

python 方法1: 设置两个指针，一个先走k-1步，然后两个同时走，当先走的到达链表尾时，另一个到达倒数第k个:

class Solution:
    def FindKthToTail(self, head, k):
        if head is None or k <= 0:
            return None
        temp1 = head
        temp2 = head
        while k > 1:
            if temp1.next is not None:
                temp1 = temp1.next
                k -= 1
            else:
                return None
        while temp1.next:
            temp1 = temp1.next
            temp2 = temp2.next
        return temp2

python 方法2：获取链表长度，从而可以得到倒数第k个节点是正数第多少个:

class Solution:
    def FindKthToTail(self, head, k):
        # 链表节点总数
        number = 0
        cur = head
        while cur:
            number += 1
            cur = cur.next
        # 倒数第k个节点在链表中正数第多少个
        index = number - k
        if index < 0:
            return None
        temp = head
        while index > 0:
            temp = temp.next
            index -= 1
        return temp

c++:

class Solution {
public:
    ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
        if(k == 0 || pListHead == NULL){
            return NULL;
        }
        ListNode *pTail = pListHead;
        ListNode *pHead = pListHead;
        // pHead 节点指针向后移动k-1步
        while(k>1){
            if(pHead->next == NULL){
                return NULL;
            }
            else{
                pHead = pHead->next;
                k--;
            }
        }
        while(pHead->next){
            pHead = pHead->next;
            pTail = pTail->next;
        }
        return pTail;
    }
};