Compare Version Numbers

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.

public class Solution {    //runtime = 249mspublic int compareVersion(String version1, String version2) {        char ch1[] = version1.toCharArray();        char ch2[] = version2.toCharArray();        int i = 0;        int j = 0;        int len1 = 0;        int len2 = 0;        int num1 = 0;        int num2 = 0;        while (i < ch1.length && (j < ch2.length)){                len1 = i;        len2 = j;        while (i < ch1.length && (ch1[i] != '.')){        i ++;        }        while (j < ch2.length && (ch2[j] != '.')){        j ++;        }        num1 = getNum(ch1, len1, i);         num2 = getNum(ch2, len2, j);        if (panDuan(num1, num2) == 0){        i ++;        j ++;        }        else {        return panDuan(num1, num2);        }        }        if (i > ch1.length && (j < ch2.length)){        return (-1) * isEqual(ch2, j, ch2.length);        }        if (i < ch1.length && (j > ch2.length)){        return isEqual(ch1, i, ch1.length);        }        return panDuan(num1, num2);    }        public int getNum(char ch[], int l, int h){    int num = 0;    while (l < h){    num += (ch[l] - 48) * Math.pow(10, h - 1 - l);    l ++;    }    return num;    }        public int panDuan(int num1, int num2){    if (num1 == num2){    return 0;    }    else {    return num1 > num2 ? 1:-1;    }    }        public int isEqual(char ch[], int i, int len){    while (i < len && (ch[i] - 48 == 0 || ch[i] == '.')){    i ++;    }    return i == len ? 0 : 1;    }}

虽然算法写是写出来了，而且运行显示Aceepted，但是实在是太冗长了！！！于是乎去discuss里找了找，发现了有位同学用Java写的解法超级好！！于是模仿着写了一个，用到了几个Java自带函数，虽然runtime也没有很短，但是真的很好用哦~~~

public class Solution {public int compareVersion(String version1, String version2){String s1[] = version1.split("\\.");String s2[] = version2.split("\\.");int len1 = s1.length;int len2 = s2.length;int i = 0;int j = 0;while ((i < len1) && (j < len2)){int d1 = Integer.parseInt(s1[i]);int d2 = Integer.parseInt(s2[i]);if (d1 > d2){return 1;}if (d1 < d2){return -1;}i ++;j ++;}while (i < len1){if (Integer.parseInt(s1[i]) > 0){return 1;}i ++;}while (j < len2){if (Integer.parseInt(s2[j]) > 0){return -1;}j ++;}return 0;}}

Integer.parseInt()把当前String对象转换成int.

StringObject.split([separator，[limit]])函数：对StringObj 中每个出现 separator 的位置进行分解，返回的是一个字符型数组。operator：字符串或正则表达式对象，它标识了分隔字符串时使用的是一个或多个字符。如果忽略该选项，则返回包含整个字符串的单一元素数组；limit，用来限制返回数组中的元素个数。