第36题 Compare Version Numbers

Compare two version numbers version1 and version1.
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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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Solution in Java:

public class Solution {    public int compareVersion(String version1, String version2) {        String[] v1 = version1.split("\\.");        String[] v2 = version2.split("\\.");        for(int i=0; i<Math.min(v1.length, v2.length); i++){            if(Integer.parseInt(v1[i])<Integer.parseInt(v2[i])) return -1;            else if(Integer.parseInt(v1[i])>Integer.parseInt(v2[i])) return 1;        }        boolean zero=true;        if(v1.length<v2.length){            for(int i=v1.length; i<v2.length; i++)                zero = zero&(Integer.parseInt(v2[i])==0);            if(!zero) return -1;        }        if(v1.length>v2.length){            for(int i=v2.length; i<v1.length; i++)                zero = zero&(Integer.parseInt(v1[i])==0);            if(!zero) return 1;        }         return 0;    }}

Note:

用“.”分割字符串时，注意需要转义成“\\.”，其中“\”自身还需要转义一次。

version“1.0” version“1”和version“1.0.0.0”是相等的，所以当两个version长度不一样时，需要检查长的一方后面几个部分是否都为0，如果是则返回0，相等。

Solution in C++:

class Solution {public:    int compareVersion(string version1, string version2) {        int ptr1 = 0, ptr2 = 0;        int val1 = 0, val2=0;        while(ptr1<version1.size()||ptr2<version2.size()){            while(ptr1<version1.size()&&version1[ptr1]!='.'){                val1 =  val1*10 + (version1[ptr1]-'0');                ptr1++;            }            while(ptr2<version2.size()&&version2[ptr2]!='.'){                val2 = val2*10 + (version2[ptr2]-'0');                ptr2++;            }            if(val1>val2) return 1;            if(val1<val2) return -1;            val1=0;            val2=0;            ptr1++;            ptr2++;        }        return 0;    }};

Note: 不关心格式，val1和val2初值赋为0，只关心该段内值的大小。