《leetCode》：Compare Version Numbers

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 . 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

思路

以点来分割字符串，并逐个将其转换为数字进行比较即可。

public  int compareVersion(String version1, String version2) {
        String[] str1=version1.split("\\.");
        String[] str2=version2.split("\\.");
        int len1=str1.length;
        int len2=str2.length;
        for(int i=0;i<min(len1,len2);i++){
            if(Integer.parseInt(str1[i])<Integer.parseInt(str2[i])){
                return -1;
            }
            else if(Integer.parseInt(str1[i])>Integer.parseInt(str2[i])){
                return 1;
            }
        }
        if(len1>len2){//检查str1 后面的字符串是否全为 0
            for(int j=len2;j<len1;j++){
                if(Integer.parseInt(str1[j])!=0){
                    return 1;
                }
            }
        }
        else if(len1<len2){
            for(int j=len1;j<len2;j++){
                if(Integer.parseInt(str2[j])!=0){
                    return -1;
                }
            }
        }
        return 0;           
    }
    private  int min(int a,int b){
        return (a<b)?a:b;
    }

用C语言实现如下：

int compareVersion(char* version1, char* version2) {
    //根据题目要求，version1和version2一定有效
    int len1=strlen(version1);
    int len2=strlen(version2) ;
    if(len1==0||strlen==0){
        return 0;
    }
    int i=0,j=0;
    int num1=0;
    int num2=0;
    while(i<len1||j<len2){
        num1=0;
        num2=0;
        while(i<len1&&version1[i]!='.'){
            num1=num1*10+version1[i]-'0';
            i++;
        }
        while(j<len2&&version2[j]!='.'){
            num2=num2*10+version2[j]-'0';
            j++;
        }
        if(num1>num2){
            return 1;
        }
        if(num1<num2){
            return -1;
        }
        if(version1[i]=='.'){
            i++;
        }
        if(version2[j]=='.'){
            j++;
        }
    }
    return 0;
}