【刷题-LeetCode】165 Compare Version Numbers

1. 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.

You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note:

1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
2. Version strings do not start or end with dots, and they will not be two consecutive dots.

解 解析出每一段版本号进行比较，假设解析出来的数字分别为n1, n2

• n1 > n2：版本1>版本2
• n1 < n2：版本1 < 版本2
• n1 == n2 ：比较下一段

在比较到末尾时，长的版本号剩余部分如果全部为0，则版本1 == 版本2，否则长的号对应的版本高

为了方便，在每个版本号后面增加了'#'作为结束符

class Solution {
public:
    int compareVersion(string version1, string version2) {
        version1.push_back('#');
        version2.push_back('#');
        int pos1 = 0, pos2 = 0;
        int num1 = 0, num2 = 0;
        while(version1[pos1] != '#' && version2[pos2] != '#'){
            num1 = 0; num2 = 0;
            if(version1[pos1] == '.')pos1++;
            while(version1[pos1] != '#' && version1[pos1] != '.'){
                num1 = num1 * 10 + (version1[pos1++] - '0');
            }
            if(version2[pos2] == '.')pos2++;
            while(version2[pos2] != '#' && version2[pos2] != '.'){
                num2 = num2 * 10 + (version2[pos2++] - '0');
            }
            if(num1 < num2)return -1;
            else if(num1 > num2)return 1;
        }
        if(version1[pos1] == '#' && version2[pos2] == '#'){
            return 0;
        }else if(version1[pos1] == '#'){
            if(allZeros(version2.substr(pos2)))return 0;
            else return -1;
        }else{
            if(allZeros(version1.substr(pos1)))return 0;
            else return 1;
        }
    }
    bool allZeros(const string &s){
        int test = 0, pos = 0;
        while(s[pos] != '#'){
            if(s[pos] == '.')pos++;
            else test += s[pos++] - '0';
        }
        if(test == 0)return true;
        else return false;
    }
};