题目:
Compare two version numbers version1 and version 2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
比较版本1和版本2,如果版本1 > 版本2,返回1,如果版本1 < 版本2,则返回-1,其他情况返回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.
假设版本字符串非空,且只包含数字和字符' . '。字符' . '不是小数点,而是用来分割数字的。
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
思路:
依次取小数点前的数字进行比较。如果一个version已经到达最后,则令比较的数字为0.
代码:
class Solution { public: int compareVersion(string version1, string version2) { int i = 0 , j = 0; //循环结束条件是i和j都达到了尾部,如果有一个先达到尾部,则当前比较的数num为0 while(i < version1.size() || j < version2.size()) { int num1 = 0 , num2 = 0; while(i < version1.size() && version1[i] != '.' ) { num1 = num1 * 10 + (version1[i] - '0'); i++; } while( j < version2.size() && version2[j] != '.') { num2 = num2 * 10 + (version2[j] - '0'); j++; } if(num1 > num2) return 1; if(num1 < num2) return -1; i++; j++; } return 0; } };