LeetCode 165. Compare Version Numbers 解题报告

165. Compare Version Numbers

My Submissions

Question

Total Accepted: 45262 Total
Submissions: 268639 Difficulty: 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

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

Show Tags

Have you met this question in a real interview?
Yes

No

Discuss

这道题其实很简单，但是难在边界条件！两次WA在边界条件上。“１.０”和“１”是相等的，“１.１”是大于“１”的，注意这两个边界条件。

public class CompareVersionNumbers {

	public static void main(String[] args) {
		System.out.println(compareVersion("1.1", "1.11"));
		System.out.println(compareVersion("1.10", "1.9"));
		System.out.println(compareVersion("1.10", "1.10"));
		System.out.println(compareVersion("1.0", "1"));
		System.out.println(compareVersion("1.1", "1"));
	}

	public static int compareVersion(String version1, String version2) {
		String[] v1 = version1.split("\\.");
		String[] v2 = version2.split("\\.");
		for (int i = 0; i < v1.length || i < v2.length; i++) {
			if (i >= v1.length)
				if (Integer.parseInt(v2[i]) == 0) continue;
				else return -1;
			if (i >= v2.length)
				if (Integer.parseInt(v1[i]) == 0) continue;
				else return 1;
			int n1 = Integer.parseInt(v1[i]);
			int n2 = Integer.parseInt(v2[i]);
			if (n1 > n2)
				return 1;
			if (n1 < n2)
				return -1;
		}
		return 0;
	}
}