165. Compare Version Numbers (M)

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本文介绍了一种比较两个版本号大小的算法，通过将版本号字符串拆分为数组或直接遍历字符串，逐个比较数字大小，实现了精确的版本号对比。文章提供了两种实现方式：使用split方法和直接遍历字符。

Compare Version Numbers (M)

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.

题意

比较两个版本号的大小。

思路

比较方便的是先以"."为分隔符将字符串拆成一个字符串数组，再挨个比较每一个字符串代表的数字大小。

当然也可以不用java内置的api实现，直接遍历字符串计算每一层级版本对应的数字。

代码实现 - split

class Solution {
    public int compareVersion(String version1, String version2) {
        // 注意正则需要用"\\."来匹配"."
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");
        int i = 0;

        while (i < v1.length || i < v2.length) {
            // 长度不够则当前对应数字为0
            int x = i < v1.length ? Integer.parseInt(v1[i]) : 0;
            int y = i < v2.length ? Integer.parseInt(v2[i]) : 0;

            if (x < y) return -1;
            if (x > y) return 1;

            i++;
        }
        
        return 0;
    }
}

代码实现 - 遍历

class Solution {
    public int compareVersion(String version1, String version2) {
        int i = 0, j = 0;
        while (i < version1.length() || j < version2.length()) {
            int x = 0, y = 0;		// 长度不够时则为默认值0
            
            while (i < version1.length() && version1.charAt(i) != '.') {
                x = x * 10 + version1.charAt(i++) - '0';
            }
            i++;		// 移动到"."后一位数字
            while (j < version2.length() && version2.charAt(j) != '.') {
                y = y * 10 + version2.charAt(j++) - '0';
            }
            j++;		// 移动到"."后一位数字

            if (x < y) return -1;
            if (x > y) return 1;
        }
        return 0;
    }
}