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Problem
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: -1Example 2:
Input: version1 = “1.0.1”, version2 = “1”Output: 1Example 3:
Input: version1 = “7.5.2.4”, version2 = “7.5.3”Output: -1Example 4:
Input: version1 = “1.01”, version2 = “1.001”Output: 0Explanation: Ignoring leading zeroes, both“01”and“001″ represent the same number“1”Example 5:
Input: version1 = “1.0”, version2 = “1.0.0”Output: 0Explanation: 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.
Solution
class Solution {
public int compareVersion(String version1, String version2) {
String[] v1 = version1.split(“\\.”);
String[] v2 = version2.split(“\\.”);
int len = Math.max(v1.length, v2.length);
for (int i = 0; i < len; i++) {
int n1 = i < v1.length ? Integer.parseInt(v1[i]) : 0;
int n2 = i < v2.length ? Integer.parseInt(v2[i]) : 0;
int ans = Integer.compare(n1, n2);
if (ans != 0) return ans;
}
return 0;
}
}
