public static void main(String[] args) {

    String version1 = "1.0";
    String version2 = "1.0.0";
    String[] version1_splits = version1.split("\\.");
    String[] version2_splits = version2.split("\\.");
    int length = version1_splits.length >= version2_splits.length ? version1_splits.length : version2_splits.length;
    int i=0;
    for(;i<length;i++){
        int version1_int = getValue(version1_splits,i);
        int version2_int = getValue(version2_splits,i);
        if(version1_int > version2_int){
            System.out.println("version1 > version2");
            break;
        }
        else if(version1_int < version2_int){
            System.out.println("version2 > version1");
            break;
        }
        else{
            if(i == length-1)
            System.out.println("version1 = version2");
        }
    }
}

private static int getValue(String[] version1_splits, int i) {
    int temp;
    try{
        temp = Integer.valueOf(version1_splits[i]);
    }
    catch(IndexOutOfBoundsException e){
        temp=0;
    }

    return temp;
}

使用Java 8 Stream替换组件中的前导零。这段代码通过了interviewbit.com上的所有测试

public int compareVersion(String A, String B) {
    List<String> strList1 = Arrays.stream(A.split("\\."))
                                           .map(s -> s.replaceAll("^0+(?!$)", ""))
                                           .collect(Collectors.toList());
    List<String> strList2 = Arrays.stream(B.split("\\."))
                                           .map(s -> s.replaceAll("^0+(?!$)", ""))
                                           .collect(Collectors.toList());
    int len1 = strList1.size();
    int len2 = strList2.size();
    int i = 0;
    while(i < len1 && i < len2){
        if (strList1.get(i).length() > strList2.get(i).length()) return 1;
        if (strList1.get(i).length() < strList2.get(i).length()) return -1;
        int result = new Long(strList1.get(i)).compareTo(new Long(strList2.get(i)));
        if (result != 0) return result;
        i++;
    }
    while (i < len1){
        if (!strList1.get(i++).equals("0")) return 1;
    }
    while (i < len2){
        if (!strList2.get(i++).equals("0")) return -1;
    }
    return 0;
}

我现在就做了，然后问自己，这对吗?因为我从来没有找到过比我的更干净的解决方案

你只需要像下面这样拆分字符串版本("1.0.0"):

userVersion.split("\\.")

那么你将得到:{"1"，"0"，"0"}

现在，用我做过的方法

isUpdateAvailable(userVersion.split("\\."), latestVersionSplit.split("\\."));

方法:

/**
 * Compare two versions
 *
 * @param userVersionSplit   - User string array with major, minor and patch version from user (exemple: {"5", "2", "70"})
 * @param latestVersionSplit - Latest string array with major, minor and patch version from api (example: {"5", "2", "71"})
 * @return true if user version is smaller than latest version
 */
public static boolean isUpdateAvailable(String[] userVersionSplit, String[] latestVersionSplit) {

    try {
        int majorUserVersion = Integer.parseInt(userVersionSplit[0]);
        int minorUserVersion = Integer.parseInt(userVersionSplit[1]);
        int patchUserVersion = Integer.parseInt(userVersionSplit[2]);

        int majorLatestVersion = Integer.parseInt(latestVersionSplit[0]);
        int minorLatestVersion = Integer.parseInt(latestVersionSplit[1]);
        int patchLatestVersion = Integer.parseInt(latestVersionSplit[2]);

        if (majorUserVersion <= majorLatestVersion) {
            if (majorUserVersion < majorLatestVersion) {
                return true;
            } else {
                if (minorUserVersion <= minorLatestVersion) {
                    if (minorUserVersion < minorLatestVersion) {
                        return true;
                    } else {
                        return patchUserVersion < patchLatestVersion;
                    }
                }
            }
        }
    } catch (Exception ignored) {
        // Will be throw only if the versions pattern is different from "x.x.x" format
        // Will return false at the end
    }

    return false;
}

等待任何反馈:)

我自己写了一个小函数。更简单地使用列表

 public static boolean checkVersionUpdate(String olderVerison, String newVersion) {
        if (olderVerison.length() == 0 || newVersion.length() == 0) {
            return false;
        }
        List<String> newVerList = Arrays.asList(newVersion.split("\\."));
        List<String> oldVerList = Arrays.asList(olderVerison.split("\\."));

        int diff = newVerList.size() - oldVerList.size();
        List<String> newList = new ArrayList<>();
        if (diff > 0) {
            newList.addAll(oldVerList);
            for (int i = 0; i < diff; i++) {
                newList.add("0");
            }
            return examineArray(newList, newVerList, diff);
        } else if (diff < 0) {
            newList.addAll(newVerList);
            for (int i = 0; i < -diff; i++) {
                newList.add("0");
            }
            return examineArray(oldVerList, newList, diff);
        } else {
            return examineArray(oldVerList, newVerList, diff);
        }

    }

    public static boolean examineArray(List<String> oldList, List<String> newList, int diff) {
        boolean newVersionGreater = false;
        for (int i = 0; i < oldList.size(); i++) {
            if (Integer.parseInt(newList.get(i)) > Integer.parseInt(oldList.get(i))) {
                newVersionGreater = true;
                break;
            } else if (Integer.parseInt(newList.get(i)) < Integer.parseInt(oldList.get(i))) {
                newVersionGreater = false;
                break;
            } else {
                newVersionGreater = diff > 0;
            }
        }

        return newVersionGreater;
    }

我创建了一个简单的实用程序，使用语义版本约定在Android平台上比较版本。所以它只适用于X.Y.Z (Major.Minor.Patch)格式的字符串，其中X、Y和Z是非负整数。你可以在我的GitHub上找到它。

方法version . compareversions (String v1, String v2)比较两个版本字符串。如果版本相等则返回0，如果版本v1在版本v2之前则返回1，如果版本v1在版本v2之后则返回-1，如果版本格式无效则返回-2。

我喜欢@Peter Lawrey的想法，我把它扩展到更远的范围:

    /**
    * Normalize string array, 
    * Appends zeros if string from the array
    * has length smaller than the maxLen.
    **/
    private String normalize(String[] split, int maxLen){
        StringBuilder sb = new StringBuilder("");
        for(String s : split) {
            for(int i = 0; i<maxLen-s.length(); i++) sb.append('0');
            sb.append(s);
        }
        return sb.toString();
    }

    /**
    * Removes trailing zeros of the form '.00.0...00'
    * (and does not remove zeros from, say, '4.1.100')
    **/
    public String removeTrailingZeros(String s){
        int i = s.length()-1;
        int k = s.length()-1;
        while(i >= 0 && (s.charAt(i) == '.' || s.charAt(i) == '0')){
          if(s.charAt(i) == '.') k = i-1;
          i--;  
        } 
        return s.substring(0,k+1);
    }

    /**
    * Compares two versions(works for alphabets too),
    * Returns 1 if v1 > v2, returns 0 if v1 == v2,
    * and returns -1 if v1 < v2.
    **/
    public int compareVersion(String v1, String v2) {

        // Uncomment below two lines if for you, say, 4.1.0 is equal to 4.1
        // v1 = removeTrailingZeros(v1);
        // v2 = removeTrailingZeros(v2);

        String[] splitv1 = v1.split("\\.");
        String[] splitv2 = v2.split("\\.");
        int maxLen = 0;
        for(String str : splitv1) maxLen = Math.max(maxLen, str.length());
        for(String str : splitv2) maxLen = Math.max(maxLen, str.length());
        int cmp = normalize(splitv1, maxLen).compareTo(normalize(splitv2, maxLen));
        return cmp > 0 ? 1 : (cmp < 0 ? -1 : 0);
    }

希望它能帮助到别人。它通过了interviewbit和leetcode中的所有测试用例(需要取消compareVersion函数中的两行注释)。

很容易测试!