LeetCode 165 Compare Version Numbers(比较版本号)(string)(*)
来源:互联网 发布:7号零食淘宝店 编辑:程序博客网 时间:2024/04/29 09:02
翻译
比较两个版本号version1和version2。
如果version1大于version2返回1,如果version1小于version2返回-1,否则返回0。
你可以假设版本号字符串是非空并且只包含数字和“.”字符。
“.”字符不代表十进制中的点,而被用作分隔数字序列。
例如,2.5不是“两个半”,也不是“差一半到三”,而是第二版中的第五个小版本。
这里有一个版本号排序的示例:
0.1 < 1.1 < 1.2 < 13.37
原文
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
分析
看到这道题的时候我还沾沾自喜,因为思路立马有了,然而……
我想到的是,将字符串根据“.”来切开,然后将数字从string转到int保存到vector中,最后比较vector里面的数据。
可是我没想到居然还有这么变态的测试用例……
"19.8.3.17.5.01.0.0.4.0.0.0.0.0.0.0.0.0.0.0.0.0.00.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.000000.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.000000""19.8.3.17.5.01.0.0.4.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0000.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.000000"
然后我就跪了……
下面是string转vector的代码,仅作为个人笔记用……
vector<int> getVersionVector(string version) { vector<int> versionV; istringstream sstream(version); string temp; while (!sstream.eof()) { getline(sstream, temp, '.'); versionV.push_back(stoi(temp)); } return versionV;}
还是去看别人的代码好了,下次再战!
class Solution {public: int compareVersion(string version1, string version2) { istringstream s1(version1), s2(version2); string a, b; while (!s1.eof() && !s2.eof()) { getline(s1, a, '.'); getline(s2, b, '.'); if (stoi(a) == stoi(b)) continue; else return stoi(a) > stoi(b) ? 1 : -1; } if (s1.eof() && s2.eof()) return 0; if (s1.eof()) { while (!s2.eof()) { getline(s2, b, '.'); if (stoi(b) != 0) return -1; } } if (s2.eof()) { while (!s1.eof()) { getline(s1, a, '.'); if (stoi(a) != 0) return 1; } } return 0; }};
class Solution {public: int compareVersion(string version1, string version2) { size_t b1 = 0, i1 = 0, b2 = 0, i2 = 0; while (i1 < version1.size() || i2 < version2.size()) { while (i1 < version1.size() && version1[i1] != '.') ++i1; while (i2 < version2.size() && version2[i2] != '.') ++i2; string sub1 = (i1 == b1) ? "0" : version1.substr(b1, i1); string sub2 = (i2 == b2) ? "0" : version2.substr(b2, i2); int ii1 = stoi(sub1), ii2 = stoi(sub2); if (ii1 > ii2) return 1; else if (ii1 < ii2) return -1; else { b1 = (b1 == i1) ? i1 : (i1++) + 1; b2 = (b2 == i2) ? i2 : (i2++) + 1; } } return 0; }};
int compareVersion(string version1, string version2) { istringstream v1(version1+"."), v2(version2+'.'); char dot = '.'; int val1 = 0, val2 = 0; while (true) { void* p1 = (v1>>val1>>dot), *p2= (v2>>val2>>dot); if (! p1 && !p2) return 0; if (! p1) val1 = 0; if (! p2) val2 = 0; if (val1>val2) return 1; else if (val1<val2) return -1; }}
updated at 2016/09/19
用Java写了一遍,其实思路就那样,没什么特别好的想法,算是暴搜吧,强行把代码折叠了一下。
public int compareVersion(String version1, String version2) { if (version1 == null || version1.length() <= 0) return 0; if (version2 == null || version2.length() <= 0) return 0; String[] vs1 = version1.split("\\."), vs2 = version2.split("\\."); int len1 = vs1.length, len2 = vs2.length; int i = 0; for (; i < len1 && i < len2; i++) { if (getInt(vs1[i]) > getInt(vs2[i])) return 1; else if (getInt(vs1[i]) < getInt(vs2[i])) return -1; } if ((len1 - len2) > 1) { int w = 0; for (int j = i; j < len1; j++) { if (getInt(vs1[j]) == 0) w++; } if (w == len1 - i) return 0; return 1; } else if ((len1 - len2) == 1) { if (getInt(vs1[i]) != 0) return 1; else return 0; } if ((len2 - len1) > 1) { int w = 0; for (int j = i; j < len2; j++) { if (getInt(vs2[j]) == 0) w++; } if (w == len2 - i) return 0; return -1; } else if ((len2 - len1) == 1) { if (getInt(vs2[i]) != 0) return -1; else return 0; } return 0; } private int getInt(String str) { return Integer.parseInt(str); }
- LeetCode 165 Compare Version Numbers(比较版本号)(string)(*)
- 165. Compare Version Numbers (版本号比较)
- 版本号比较(Compare two Version numbers)
- 【LeetCode-面试算法经典-Java实现】【165-Compare Version Numbers(比较版本号)】
- leetcode Compare Version Numbers版本号比较
- (Leetcode) Compare Version Numbers --- 比较版本号
- leetcode 165. Compare Version Numbers 版本号比较
- LeetCode(165) Compare Version Numbers
- LeetCode OJ 之 Compare Version Numbers (比较版本数)
- LeetCode 165. Compare Version Numbers(比较版本)
- LeetCode(68)-Compare Version Numbers
- LeetCode[String]: Compare Version Numbers
- [leetcode 165] Compare Version Numbers
- leetcode #165 Compare Version Numbers
- leetcode-165 Compare Version Numbers
- [LeetCode 165]Compare Version Numbers
- [LeetCode][165][Compare Version Numbers]
- Compare Version Numbers- LeetCode 165
- java Runtime.getRuntime().exec 调用系统脚本/命令注意事项
- rate limiting
- Jetty多项目启动时的冲突问题
- jQuery - jQuery的$.extend和$.fn.extend作用及区别
- Latex入门
- LeetCode 165 Compare Version Numbers(比较版本号)(string)(*)
- android.hardware.camera2使用指南
- hdu2795 Billboard
- MyBatis Generator 详解
- 单机系统和分布式系统相关概念对比
- 启动Tomcat报异常host-manager does not exist or is not a readable directory
- Python 3.x基于Xml数据的Http请求
- UINavigationController 弹出新的UIViewController时,setNavigationBarHidden失效的问题
- 用C++进行hadoop程序开发(hadoop Pipes)