Compare Version Numbers.java

M
1521777505
tags: String

给两串version number, 由数字和'.' 组成. 比较先后顺序. 

If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

#### divide and conquer 
- 用 str.split("\\.") 分割string
- Convert成integer, 逐个击破

#### 注意
- '1.0' 和 '0' 是相等的
- 如果可以假设version integer都是valid, 直接Integer.parseInt()就可以了
- 不然的话, 可以compare string

```
/*
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
*/

/*
Thoughts:
- divide and conqure: divide by '.'
- during the same cun, the one with no sub-division, it's the leading version
- use Integer.parseInt() with assumption that the numbers are valid integer.
*/
class Solution {
 public int compareVersion(String version1, String version2) {
 if (version1.equals(version2)) {
 return 0;
 }
 String[] subVersion1 = version1.split("\\.");
 String[] subVersion2 = version2.split("\\.");
 int size = Math.max(subVersion1.length, subVersion2.length);
 int rst = 0;
 for (int i = 0; i < size; i++) {
 if (i >= subVersion1.length) {
 rst = Integer.parseInt(subVersion2[i]) == 0 ? 0 : -1; // assume the missing subVersion1 = 0
 } else if (i >= subVersion2.length) {
 rst = Integer.parseInt(subVersion1[i]) == 0 ? 0 : 1; // assume the missing subVersion2 = 0
 } else { // both exist
 if (Integer.parseInt(subVersion1[i]) != Integer.parseInt(subVersion2[i])) {
 rst = Integer.parseInt(subVersion1[i]) < Integer.parseInt(subVersion2[i]) ? -1 : 1; 
 }
 }
 if (rst != 0) {
 return rst;
 }
 }
 return rst;
 }
}


/*
Thoughts:
- divide and conqure: divide by '.'
- during the same cun, the one with no sub-division, it's the leading version
- compare string
*/
class Solution {
 public int compareVersion(String version1, String version2) {
 if (version1.equals(version2)) {
 return 0;
 }
 if (!version1.contains(".") && !version2.contains(".")) {
 return compareString(version1, version2);
 }
 String[] subVersion1 = version1.split("\\.");
 String[] subVersion2 = version2.split("\\.");
 int size = Math.max(subVersion1.length, subVersion2.length);
 int rst = 0;
 for (int i = 0; i < size; i++) {
 if (i >= subVersion1.length) {
 rst = compareString("0", subVersion2[i]); 
 } else if (i >= subVersion2.length) {
 rst = compareString(subVersion1[i], "0");
 } else {
 rst = compareString(subVersion1[i], subVersion2[i]); 
 }
 if (rst != 0) {
 return rst;
 }
 }
 return rst;
 }
 
 /*
 Assume the number can be super large, and can't be saved in Integer, or Long.
 Compare number by each digit
 */
 private int compareString(String str1, String str2) {
 if (str1.equals(str2)) {
 return 0;
 }
 while (str1 != null && str1.length() > 1 && str1.charAt(0) == '0') {
 str1 = str1.substring(1);
 }
 
 while (str2 != null && str2.length() > 1 && str2.charAt(0) == '0') {
 str2 = str2.substring(1);
 }
 
 if (str1.length() != str2.length()) {
 return str1.length() < str2.length() ? -1 : 1;
 }
 for (int i = 0; i < str1.length(); i++) {
 int digit1 = str1.charAt(i) - '0';
 int digit2 = str2.charAt(i) - '0';
 if (digit1 < digit2) {
 return -1;
 } else if (digit1 > digit2) {
 return 1;
 }
 }
 return 0;
 }
}
```