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/

题解：

比较版本大小。这道题目可以把输入两个string split一下，注意因为"."是特殊regex，所以要用"\\."。Split完毕以后比较array每个对应元素的大小就可以了。

Time Complexity - O(n)， Splace Complexity - O(n)。

public class Solution {
    public int compareVersion(String version1, String version2) {
        if(version1 == null || version2 == null)
            return 0;
        String[] version1Arr = version1.split("\\.");
        String[] version2Arr = version2.split("\\.");
        int i = 0, j = 0;
        
        while(i < version1Arr.length || j < version2Arr.length) {
            int ver1 = i < version1Arr.length ? Integer.parseInt(version1Arr[i]) : 0;
            int ver2 = j < version2Arr.length ? Integer.parseInt(version2Arr[i]) : 0;
            if(ver1 < ver2)
                return -1;
            else if(ver1 > ver2)
                return 1;
            i++;
            j++;
        }
        
        return 0;
    }
}

 

假如不想用split，那么可以节约space complexity。这里我们不用parseInt，要使用手动计算每个level version的数字，然后加以比较。

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {
    public int compareVersion(String version1, String version2) {
        if(version1 == null || version2 == null)
            return 0;
            
        int ver1 = 0, ver2 = 0, i = 0, j = 0;
        while(i < version1.length() || j < version2.length()){
            while(i < version1.length() && version1.charAt(i) != '.'){
                ver1 = ver1 * 10 + version1.charAt(i) - '0';
                i++;
            }
            
            while(j < version2.length() && version2.charAt(j) != '.' ){
                ver2 = ver2 * 10 + version2.charAt(j) - '0';
                j++;
            }
            
            if(ver1 < ver2)
                return -1;
            else if(ver1 > ver2)
                return 1;
            else{
                ver1 = 0;
                ver2 = 0;
                i++;
                j++;
            }
        }
        return 0;
    }
}

 

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二刷:

方法跟一刷一样

Java:

Time Complexity - O(n)， Splace Complexity - O(n)。

public class Solution {
    public int compareVersion(String version1, String version2) {
        if (version1 == null || version2 == null) {
            return 0;
        }
        String[] ver1Arr = version1.split("\\.");
        String[] ver2Arr = version2.split("\\.");
        int len1 = ver1Arr.length, len2 = ver2Arr.length, lo1 = 0, lo2 = 0;
        while (lo1 < len1 || lo2 < len2) {
            int ver1 = lo1 < len1 ? Integer.parseInt(ver1Arr[lo1]) : 0;
            int ver2 = lo2 < len2 ? Integer.parseInt(ver2Arr[lo2]) : 0;
            if (ver1 < ver2) {
                return -1;
            } else if (ver1 > ver2) {
                return 1;
            } else {
                lo1++;
                lo2++;
            }
        }
        return 0;
    }
}

 

不用split

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {
    public int compareVersion(String version1, String version2) {
        if (version1 == null || version2 == null) {
            return 0;
        }
        int len1 = version1.length(), len2 = version2.length(), lo1 = 0, lo2 = 0, ver1 = 0, ver2 = 0;
        while (lo1 < len1 || lo2 < len2) {
            while (lo1 < len1 && version1.charAt(lo1) != '.') {
                ver1 = 10 * ver1 + version1.charAt(lo1) - '0';
                lo1++;
            }
            while (lo2 < len2 && version2.charAt(lo2) != '.') {
                ver2 = 10 * ver2 + version2.charAt(lo2) - '0';
                lo2++;
            }
            if (ver1 < ver2) {
                return -1;
            } else if (ver1 > ver2) {
                return 1;
            } else {
                lo1++;
                lo2++;
                ver1 = 0;
                ver2 = 0;
            }
        }
        return 0;
    }
}

 

三刷:

Java:

public class Solution {
    public int compareVersion(String version1, String version2) {
        if (version1 == null || version2 == null) return 0;
        String[] v1s = version1.split("\\.");
        String[] v2s = version2.split("\\.");
        int i = 0, j = 0, res = 0;
        while (i < v1s.length || j < v2s.length) {
            int ver1 = i < v1s.length ? Integer.valueOf(v1s[i++]) : 0;
            int ver2 = j < v2s.length ? Integer.valueOf(v2s[j++]) : 0;
            if (ver1 < ver2) return -1;
            else if (ver1 > ver2) return 1;
        }
        return 0;
    }
}

 

 

Reference：

http://www.fromdev.com/2009/10/playing-with-java-string-split-basics.html