题目: Compare two version numbers version1 and version2. If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character. The . ch…

题目 Compare two version numbers version1 and version2. If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character. The . cha…

[leetcode 字符串处理]Compare Version Numbers @author:wepon @blog:http://blog.csdn.net/u012162613 1.题目 Compare two version numbers version1 and version1. If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume…

题目 Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified. You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad…

题目: You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -&…

#-*- coding: UTF-8 -*-class Solution(object):    def compareVersion(self, version1, version2):        """        :type version1: str        :type version2: str        :rtype: int        """        versionl1=version1.split('.'…

Hard! 题目描述: 给定一个单词数组和一个长度 maxWidth,重新排版单词,使其成为每行恰好有 maxWidth 个字符,且左右两端对齐的文本. 你应该使用“贪心算法”来放置给定的单词:也就是说,尽可能多地往每行中放置单词.必要时可用空格 ' ' 填充,使得每行恰好有 maxWidth 个字符. 要求尽可能均匀分配单词间的空格数量.如果某一行单词间的空格不能均匀分配,则左侧放置的空格数要多于右侧的空格数. 文本的最后一行应为左对齐,且单词之间不插入额外的空格. 说明: 单词是指由非空格字…

[LeetCode]165. Compare Version Numbers 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/compare-version-numbers/description/ 题目描述: Compare two version numbers version1 and vers…

Compare Version Numbers Compare two version numbers version1 and version2. If *version1* > *version2* return 1; if *version1* < *version2* return -1;otherwise return 0. You may assume that the version strings are non-empty and contain only digits an…

Question 165. Compare Version Numbers Solution 题目大意: 比较版本号大小 思路: 根据逗号将版本号字符串转成数组,再比较每个数的大小 Java实现: public int compareVersion(String version1, String version2) { String[] v1Arr = version1.split("\\."); String[] v2Arr = version2.split("\\.&qu…