[LeetCode]165. Compare Version Numbers

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本文介绍了一种用于比较软件版本号的高效算法，并提供了两种实现方式：一种使用字符串分割的方法，另一种采用逐字符解析的方式。这两种方法均能准确判断出两个版本号之间的大小关系。

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Easy

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

3ms:

public int compareVersion(String version1, String version2) {
    String[] v1 = version1.split("\\.");
    String[] v2 = version2.split("\\.");
    int len1 = v1.length;
    int len2 = v2.length;
    for(int i=0;i<Math.min(len2, len1 );i++){
        int vv1 =Integer.parseInt(v1[i]);
        int vv2 = Integer.parseInt(v2[i]);
        if(vv1>vv2) return 1;
        if(vv1<vv2) return -1;
    }
    if(len1>len2){
        int t=1;
        int sub = Integer.parseInt(v1[len2]);
        while(sub==0&&len1>(len2+t)){
            sub = Integer.parseInt(v1[len2+t]);
            t++;
        }
        return sub>0?1:0;
    }
    if(len2>len1){
        int t=1;
        int sub = Integer.parseInt(v2[len1]);
        while(sub==0&&len2>(len1+t)){
            sub = Integer.parseInt(v2[len1+t]);
            t++;
        }
        return sub>0?-1:0;
    }
    return 0;
}

1ms:

public int compareVersion(String version1, String version2) {
    int i = 0, j = 0;
    int len1 = version1.length(), len2 = version2.length();
    char[] c1 = version1.toCharArray();
    char[] c2 = version2.toCharArray();
    while (i < len1 || j < len2) {
        int cur1 = 0, cur2 = 0;
        while (i < len1 && c1[i] != '.') 
            cur1 = cur1 * 10 + c1[i++] - '0';
        while (j < len2 && c2[j] != '.') 
            cur2 = cur2 * 10 + c2[j++] - '0';
        if (cur1 > cur2) return 1;
        if (cur1 < cur2) return -1;
        i++;
        j++;
    }
    return 0;
}