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

谁践踏了优雅 2022-05-19 05:45 75阅读 0赞

# 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 json
    
    class ListNode:
        def __init__(self, x):
            self.val = x
            self.next = None
    
    class Solution:
        def removeNthFromEnd(self, head, n):
            """ :type head: ListNode :type n: int :rtype: ListNode """
            res = ListNode(None)
            res.next = head
            cur = runner = res
    
            for _ in range(n):
                runner = runner.next
    
            while runner.next is not None:
                cur = cur.next
                runner = runner.next
    
            cur.next = cur.next.next
    
            return res.next
    
    def stringToIntegerList(input):
        return json.loads(input)
    
    def stringToListNode(input):
        # Generate list from the input
        numbers = stringToIntegerList(input)
        # Now convert that list into linked list
        dummyRoot = ListNode(0)
        ptr = dummyRoot
        for number in numbers:
            ptr.next = ListNode(number)
            ptr = ptr.next
    
        ptr = dummyRoot.next
        return ptr
    
    def listNodeToString(node):
        if not node:
            return "[]"
        result = ""
        while node:
            result += str(node.val) + ", "
            node = node.next
        return "[" + result[:-2] + "]"
    
    def main():
        import sys
        def 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:
                break
    
    if __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