剑指offer 面试题22（链表中倒数第K个节点） python

题目描述

输入一个链表，输出该链表中倒数第k个结点。

运行时间：28ms

占用内存：5728k

方法一：

# -*- coding:utf-8 -*-

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        # 方法一：遍历两次，第一次遍历算出链表有多少个节点，并算出倒数第K个是顺数第几个。第二个遍历就是找出该节点。
        p = head
        n = 0
        while p is not None:
            n += 1
            p = p.next
        if k > n:
            return
        else:
            m = n - k + 1
            p = head
            for i in range(1,m):
                p = p.next
            return p

方法二：

运行时间：21ms

占用内存：5728k

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        # 方法二：遍历一次得到答案。但是需要用到两个指针。第一个指针，先走k-1步，然后第k步的时候，第二个指针也开始走
        # 那么当第一个指针到达尾部的时候也是第二个指针到达倒数第K个的时候。
        if k == 0:
            return
        p1 = head
        p2 = head
        n = 1
        while p1 is not None:
            if n <= k-1:
                # 先判断会不会长度会不会大于k，如果小于k那么就直接return false
                if p1.next is not None:
                    p1 = p1.next
                    n += 1
                else:
                    return
            else:
                if p1.next is not None:
                    p2 = p2.next
                    p1 = p1.next
                else:
                    return p2