算法-找出链表的倒数第K个节点

1.题目描述

输入一个链表，输出该链表中倒数第k个节点。为了符合大多数人的习惯，本题从1开始计数，即链表的尾节点是倒数第1个节点

示例 1：
输入：head = [1,2,3,4,5], n = 2
输出：4

难度: 简单

2.解法

方法一： 常规解法

思路：
1.先遍历统计链表长度，记为N
2.再次遍历链表，N-K即为倒数第K个节点

class Node():
    def __init__(self, value, next=None):
        self.value = value
        self.next = next


class LinkNode():
    def __init__(self):
        self.head = None
        self.len = 0

    def appendNode(self, value):
        newNode = Node(value)
        if not self.head:
            self.head = newNode
            self.len += 1
        else:
            tmp = self.head
            while tmp.next:
                tmp = tmp.next
            tmp.next = newNode
            self.len += 1

    def print_link(self):
        if not self.head:
            return
        temp = self.head
        while temp.next:
            print(temp.data)
            temp = temp.next


class Solution():
    # 暴力解法
    def getKthFromEnd(self, head, k):
        count = 0
        current_node, new_node = head, head

        while current_node:
            current_node = current_node.next
            count += 1
        i = 0
        while i <= count - k:
            new_node = new_node.next
            i += 1
        return new_node


if __name__ == '__main__':
    linkNode = LinkNode()
    for i in range(8):
        linkNode.appendNode(i)
    solution = Solution()
    res=solution.getKthFromEnd(linkNode.head, 2)
    print(res.value)

时间复杂度O(N) 空间复杂度O(1)

方法二：双指针

思路：
1.初始化双指针former和latter,指针former先于latter往前走K步
2.former和latter同时往前移动，直到former到达链表末尾,此时latter即为倒数第K个节点

class Solution():
    def getKthFormEndNew(self, head, k):
        former, latter = head, head
        for _ in range(k):
            former = former.next
        while former:
            former, latter = former.next, latter.next
        return latter


if __name__ == '__main__':
    linkNode = LinkNode()
    for i in range(8):
        linkNode.appendNode(i)
    solution = Solution()
    res=solution.getKthFormEndNew(linkNode.head, 2)
    print(res.value)

时间复杂度O(N) 空间复杂度O(1)