leetcode165 - Compare Version Numbers - medium

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.
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

 

字符串处理。
1.split(“\\.”)
2.按两者更长的数组遍历过去。如果其中一个取不到数字了，就取0作为当前的数。（即隐形把1.0.1, 1.0的比较转为1.0.1, 1.0.0的比较）。

细节：
1.split(“.”)中的.是字符串匹配的任意单个字符的那张百搭牌！如果你想要声明literally的.来分割，比如这题，就要用split(“\\.”)。
2.注意corner case: 1.0和1是相等的！所以你不能只比到两者公共长度就结束了，后面的还得看是不是都为0. 但先比公共part再看后面是否为0的写法太不优雅了（参见我的辣鸡写法）。请换个思路，对比两者中更长的长度，如果取不到则补0接着对比的写法（参见leetcode摘抄写法）。

 

我的实现：

class Solution {
    public int compareVersion(String version1, String version2) {
        
        // P1: 注意.的分隔符
        String[] nums1 = version1.split("\\.");
        String[] nums2 = version2.split("\\.");
        
        int i = 0;
        while (i < nums1.length && i < nums2.length) {
            int n1 = Integer.parseInt(nums1[i]);
            int n2 = Integer.parseInt(nums2[i]);
            if (n1 < n2) {
                return -1;
            } else if (n1 > n2) {
                return 1;
            } else {
                i++;
            }
        }

        if (nums1.length == nums2.length) {
            return 0;
        }
        // P2: case 1.0和1是相等的！
        while (i < nums2.length) {
            if (Integer.parseInt(nums2[i++]) != 0) {
                return -1;
            }
        }
        while (i < nums1.length) {
            if (Integer.parseInt(nums1[i++]) != 0) {
                return 1;
            }
        }
        return 0;
    }
}

 

leetcode陈独秀网友实现：

public int compareVersion(String version1, String version2) {
    String[] levels1 = version1.split("\\.");
    String[] levels2 = version2.split("\\.");
    
    int length = Math.max(levels1.length, levels2.length);
    for (int i=0; i<length; i++) {
        Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0;
        Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0;
        int compare = v1.compareTo(v2);
        if (compare != 0) {
            return compare;
        }
    }
    
    return 0;
}