165 Compare Version Numbers

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.
You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

Example：

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note：

Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
Version strings do not start or end with dots, and they will not be two consecutive dots.

解释下题目：

给定两个版本号的字符串，比较它们的大小

1. 分割然后比较呗

实际耗时：1ms

public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");
        int len = Math.min(v1.length, v2.length);

        for (int i = 0; i < len; i++) {
            int a = Integer.valueOf(v1[i]);
            int b = Integer.valueOf(v2[i]);
            if (a > b) {
                return 1;
            } else if (a < b) {
                return -1;
            }
        }
        if (v1.length == v2.length) {
            return 0;
        } else if (v1.length < v2.length) {
            for (int i = v1.length; i < v2.length; i++) {
                int a = Integer.valueOf(v2[i]);
                if (a > 0) {
                    return -1;
                }
            }
            return 0;
        } else {
            for (int i = v2.length; i < v1.length; i++) {
                int a = Integer.valueOf(v1[i]);
                if (a > 0) {
                    return 1;
                }
            }
            return 0;
        }
    }

  思路其实很简单，分割比较呗。这里我感觉要注意的知识点有两个：

• 一个是点号作为分割符的话需要两个转义字符，即String[] v1 = version1.split("\\.");
• 还有一个是Integer.valueOf()和Integer.parseInt()这两个函数，它们都可以以字符串作为参数，不同的是value这个返回Integer对象，而parseInt这个返回的是基本类型int，所以如果了解包装类的话，这个就很好理解了。

时间复杂度O(n) n为较长的那个字符串的长度

空间复杂度O(1)