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LeetCode

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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: -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.

Code

1.12.1 > 1.3.1

  1. split 的工作原理是利用正则表达式,而在正则表达式中, “.”有特殊意思,所以匹配”.”时要用转义字符”",所以在正则表达式中匹配”.”的表达式是”.”, 而在 Java 中,\又是特殊字符, 所以还要进行转义, 所以最终变成”\.”
  2. 也是特殊字符，要使用|
  3. 字符串中不能出现\，如果要使用\，需要变成\，例如 C:\home\text.txt 如果需要对\使用 split，需要用 split(“\\”)
class Solution {
    public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");

        for(int i = 0; i < Math.max(v1.length, v2.length); i++){
            int num1 = i < v1.length ? Integer.parseInt(v1[i]) : 0;
            int num2 = i < v2.length ? Integer.parseInt(v2[i]) : 0;

            if(num1 < num2) {
                return -1;
            } else if(num1 > num2){
                return 1;
            }
        }
        return 0;
    }
}