LeetCode 165 Compare Version Numbers

题目描述

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:

[code]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

分析

比较字符串版本号大小。

把字符串按照’.’分隔成字符串数组v1，v2

因为字符串数组v1，v2的长度可能不一样，要前端对齐，较短的字符串数组变长，后面为null

分别对每一个数进行比较

本来以为这道题会有，字符串转成int溢出的问题，结果没考虑这个问题也AC了~ ^_^

代码

[code]    public static int compareVersion(String version1, String version2) {

        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");

        int m = Math.max(v1.length, v2.length);

        v1 = Arrays.copyOf(v1, m);
        v2 = Arrays.copyOf(v2, m);

        for (int i = 0; i < m; i++) {

            int n1 = 0;
            int n2 = 0;

            if (v1[i] != null) {
                n1 = Integer.valueOf(v1[i]);
            }

            if (v2[i] != null) {
                n2 = Integer.valueOf(v2[i]);
            }

            if (n1 < n2) {
                return -1;
            } else if (n1 > n2) {
                return 1;
            }
        }

        return 0;
    }