前言

本文是LeetCode19. Remove Nth Node From End of List解法，这个题目需要删除链表中的倒数第n个位置的元素

代码

# -*- coding: utf-8 -*-# !/usr/bin/env python# Time: 2018/6/27 23:44# Author: sty# File: 19. Remove Nth Node From End of List.py
import jsonclass ListNode:def __init__(self, x):self.val = xself.next = Noneclass Solution:def removeNthFromEnd(self, head, n):""":type head: ListNode:type n: int:rtype: ListNode"""res = ListNode(None)res.next = headcur = runner = resfor _ in range(n):runner = runner.nextwhile runner.next is not None:cur = cur.nextrunner = runner.nextcur.next = cur.next.nextreturn res.nextdef stringToIntegerList(input):return json.loads(input)def stringToListNode(input):# Generate list from the inputnumbers = stringToIntegerList(input)# Now convert that list into linked listdummyRoot = ListNode(0)ptr = dummyRootfor number in numbers:ptr.next = ListNode(number)ptr = ptr.nextptr = dummyRoot.nextreturn ptrdef listNodeToString(node):if not node:return "[]"result = ""while node:result += str(node.val) + ", "node = node.nextreturn "[" + result[:-2] + "]"def main():import sysdef readlines():for line in sys.stdin:yield line.strip('\n')lines = readlines()while True:try:line = next(lines)head = stringToListNode(line);line = next(lines)n = int(line);ret = Solution().removeNthFromEnd(head, n)out = listNodeToString(ret);print(out)except StopIteration:breakif __name__ == '__main__':main()

输入格式

[2,4,5,6,4,5,6,6]
4
[2, 4, 5, 6, 5, 6, 6]

注意

它这里使用的输入方式是列表的形式输入的

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LeetCode19. Remove Nth Node From End of List 删除链表中的倒数第n个位置的元素

前言

代码

输入格式

注意

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