共计 1850 个字符，预计需要花费 5 分钟才能阅读完成。
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 1:
Input: version1 = “0.1”, version2 = “1.1”
Output: -1
Example 2:
Input: version1 = “1.0.1”, version2 = “1”
Output: 1
Example 3:
Input: version1 = “7.5.2.4”, version2 = “7.5.3”
Output: -1
Example 4:
Input: version1 = “1.01”, version2 = “1.001”
Output: 0
Explanation: Ignoring leading zeroes, both“01”and“001″ represent the same number“1”
Example 5:
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.
难度：medium
题目：比较两个版本 version1 和 version2. 如果 1 大于 2 返回 1,1 小于 2 返回-1，否则返回 0. 假定版本字符串中非空且只含有数字和点。点不是小数点只是用作分隔符。
思路：按点号分割转成数字比较。
Runtime: 1 ms, faster than 89.87% of Java online submissions for Compare Version Numbers.Memory Usage: 33.1 MB, less than 100.00% of Java online submissions for Compare Version Numbers.
class Solution {
public int compareVersion(String version1, String version2) {
String[] vs1 = version1.split(“\\.”);
String[] vs2 = version2.split(“\\.”);
for (int i = 0; i < Math.max(vs1.length, vs2.length); i++) {
int i1 = i < vs1.length ? Integer.parseInt(vs1[i]) : 0;
int i2 = i < vs2.length ? Integer.parseInt(vs2[i]) : 0;
if (i1 – i2 > 0) {
return 1;
}
if (i1 – i2 < 0) {
return -1;
}
}
return 0;
}
}
