165. Compare Version Numbers**-CFANZ编程社区

165. Compare Version Numbers**

https://leetcode.com/problems/compare-version-numbers/

题目描述

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:

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

C++ 实现 1

对 version number 按照 . 进行 split (要是用 Python 的话, 直接 version.split('.') 就特别方便了), 这个功能可以使用 getline 实现, 在 71. Simplify Path** 我就采用过. 使用 std::stoi() 可以将字符串转换为整数, 该函数可以处理函数 leading zero 的字符串, 比如说可以传入 "001" 这样的字符串.

class Solution {
public:
    int compareVersion(string version1, string version2) {
        stringstream ss1(version1), ss2(version2);
        string u;
        vector<string> num1, num2;
        while (getline(ss1, u, '.')) num1.push_back(u);
        while (getline(ss2, u, '.')) num2.push_back(u);
        int i = 0, j = 0;
        while (i < num1.size() || j < num2.size()) {
            auto n1 = i >= num1.size() ? 0 : std::stoi(num1[i]);
            auto n2 = j >= num2.size() ? 0 : std::stoi(num2[j]);
            if (n1 > n2) return 1;
            else if (n1 < n2) return -1;
            ++ i, ++ j;
        }
        return 0;
    }
};

C++ 实现 2

不采用 stringstream:

class Solution {
public:
    // 注意可能有 "01.1.1" 这样的情况, 所以里面还有两个 while 循环用于处理
    // "01" 连续几个数的情况.
    int compareVersion(string version1, string version2) {
        int ver1 = 0, ver2 = 0; // 用于保存每个 level 的 version
        int i = 0, j = 0;
        while (i < version1.size() || j < version2.size()) {
            // 两个 while 循环用于求每个 level 的 version
            while (isdigit(version1[i]) && i < version1.size())
                ver1 = ver1 * 10 + version1[i++] - '0';
            
            while (isdigit(version2[j]) && j < version2.size())
                ver2 = ver2 * 10 + version2[j++] - '0';
            
            // 比较每个 level 的 version
            if (ver1 < ver2)
                return -1;
            else if (ver1 > ver2)
                return 1;
            
            ver1 = 0, ver2 = 0; // 用于下一个 level 的比较
            // 前面两个 while 循环中 i 和 j 已经指向了 '.',
            // 这里将 '.' 跳过.
            i ++;
            j ++;
        }
        // 只有每个 level 的 version 都相等, 才返回 0;
        return 0;
    }
};