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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
注意可能会出现前置0的情况，所以分为小数点钱与小数点后转换成数字来判断，代码如下：
1 class Solution { 2 public: 3 int compareVersion(string version1, string version2) { 4 int i1, i2; 5 int val1, val2; 6 for(i1 = 0, i2 = 0; i1 < version1.size() || i2 < version2.size(); ++i1, ++i2){ 7 val1 = 0; 8 for(; i1 < version1.size(); ++i1){ 9 if(version1[i1] == ‘.‘) 10 break; 11 val1 = val1 * 10 + version1[i1] - ‘0‘; 12 } 13 val2 = 0; 14 for(; i2 < version2.size(); ++i2){ 15 if(version2[i2] == ‘.‘) 16 break; 17 val2 = val2 * 10 + version2[i2] - ‘0‘; 18 } 19 if(val1 > val2) 20 return 1; 21 else if(val1 < val2) 22 return -1; 23 } 24 return 0; 25 } 26 };
LeetCode OJ：Compare Version Numbers（比较版本字符串）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4937315.html
