LeetCode（68）-Compare Version Numbers

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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-levelrevision of the second first-level revision.Here is an example of version numbers ordering:0.1 < 1.1 < 1.2 < 13.37

思路：

题意：比较两个版本号字符串的大小
把字符串用split转化为数组，注意split（\.），然后转化为整数数组，遍历比较。注意如果版本号后面都是零的情况

代码：

public class Solution {public int compareVersion(String version1, String version2) {String[] v1,v2;if(version1.indexOf(".") == -1){v1 = new String[1];v1[0] = version1;}else{v1 = new String[version1.split("\\.").length];v1 = version1.split("\\.");}if(version2.indexOf(".") == -1){v2 = new String[1];v2[0] = version2;}else{v2 = new String[version2.split("\\.").length];v2 = version2.split("\\.");}int[] array1 = sToInt(v1);int[] array2 = sToInt(v2);int nn = Math.min(array1.length,array2.length);for(int a = 0;a < nn;a++){if(array1[a] > array2[a]){return 1;}else if(array1[a] < array2[a]){return -1;}}if(array1.length > array2.length){for(int k = nn; k < array1.length;k++){if(array1[k] != 0){return 1;}}return 0;}else if(array1.length < array2.length){for(int m = nn;m < array2.length;m++){if(array2[m] != 0){return -1;}}return 0;}return 0;}public int[] sToInt(String[] ss){int n = ss.length;int[] result = new int[n];for(int i = 0;i < n;i++){try{result[i] = Integer.parseInt(ss[i]);}catch(Exception e){}}return result;}
}