165. Compare Version Numbers

题目：

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

链接： http://leetcode.com/problems/compare-version-numbers/

6/16/2017

3ms, 21%

注意的问题

1. 第6，7行，string.split()对于以dot分隔的话，需要转义成\.

2. Integer.parseInt(), Integer.toString()分别用于string->int, int->string

3. 前一部分每一块需要比较empty string

4. 如果长度不等的话，不但需要比较长度，还要比较多余长度里面是不是都是0，如果是的话2个version还是一样的。

public class Solution {
    public int compareVersion(String version1, String version2) {
        if (version1 == null || version2 == null) {
            return 0;
        }
        String[] v1 = version1.split("\.");
        String[] v2 = version2.split("\.");
        
        for (int i = 0, j = 0; i < v1.length && j < v2.length; i++, j++) {
            if (v1[i].equals("") && v2[i].equals("")) {
                continue;
            }
            if (v1[i].equals("") || Integer.parseInt(v1[i]) < Integer.parseInt(v2[i])) {
                return -1;
            } else if (v2[i].equals("") || Integer.parseInt(v2[i]) < Integer.parseInt(v1[i])) {
                return 1;
            }
        }
        if (v1.length > v2.length) {
            for (int i = v2.length; i < v1.length; i++) {
                if (Integer.parseInt(v1[i]) != 0) {
                    return 1;
                }
            }
            return 0;
        } else if (v1.length < v2.length) {
            for (int i = v1.length; i < v2.length; i++) {
                if (Integer.parseInt(v2[i]) != 0) {
                    return -1;
                }
            }
            return 0;            
        }
        return 0;
    }
}

别人的答案，2个版本都比我的好，一个是将empty string变为0，一个直接用数字来计算。

http://www.cnblogs.com/yrbbest/p/4491633.html

其他答案

https://discuss.leetcode.com/topic/6238/accepted-small-java-solution

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;
}

https://discuss.leetcode.com/topic/11410/my-2ms-easy-solution-with-c-c

int compareVersion(string version1, string version2) {
    int i = 0; 
    int j = 0;
    int n1 = version1.size(); 
    int n2 = version2.size();
    
    int num1 = 0;
    int num2 = 0;
    while(i<n1 || j<n2)
    {
        while(i<n1 && version1[i]!='.'){
            num1 = num1*10+(version1[i]-'0');
            i++;
        }
        
        while(j<n2 && version2[j]!='.'){
            num2 = num2*10+(version2[j]-'0');;
            j++;
        }
        
        if(num1>num2) return 1;
        else if(num1 < num2) return -1;
        
        num1 = 0;
        num2 = 0;
        i++;
        j++;
    }
    
    return 0;
}

更多讨论

https://discuss.leetcode.com/category/173/compare-version-numbers