zoukankan      html  css  js  c++  java
  • [LeetCode] 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 and the . character.
    The . character does not represent a decimal point and is used to separate number sequences.
    For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

    Here is an example of version numbers ordering:

    0.1 < 1.1 < 1.2 < 13.37


    该题的主要意图在于考察思考问题的全面性,该题体现了C/C++的性能优势,java代码贴出.

    public class Solution {
        public int compareVersion(String version1, String version2) {
            String[] sv1 = version1.split("\.");
            String[] sv2 = version2.split("\.");
            int idx1 = 0;
            int idx2 = 0;
            
            while(idx1 < sv1.length || idx2 < sv2.length){
                int v1 = 0;
                if(idx1 < sv1.length){
                    v1 = Integer.parseInt(sv1[idx1]);
                    idx1++;
                }
                int v2 = 0;
                if(idx2 < sv2.length){
                    v2 = Integer.parseInt(sv2[idx2]);
                    idx2++;
                }
                if(v1 < v2)
                    return -1;
                if(v1 > v2)
                    return 1;
            }
            return 0;
        }
    }
  • 相关阅读:
    python中的内置函数的思维导图
    练习(面试题):关于生成器函数的求和问题
    推导式, 生成器表达式
    生成器
    静态代码块
    java中内存的划分
    静态方法
    Chapter07Scanner类、Random类、ArrayList类
    泛型
    对象数组
  • 原文地址:https://www.cnblogs.com/HackingProgramer/p/4354757.html
Copyright © 2011-2022 走看看