#Leetcode# 165. Compare Version Numbers-优快云博客

本文详细解析了如何通过算法比较两个版本号的大小，通过分解版本号字符串并将其转换为整数序列进行逐级对比，提供了完整的C++实现代码，并讨论了处理前导零和不同长度版本号的策略。

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https://leetcode.com/problems/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.

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

代码：

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int len1 = version1.size(), len2 = version2.size();
        int num1[100100], num2[100100];
        int cnt1 = 0, cnt2 = 0;
        memset(num1, 0, sizeof(num1));
        memset(num2, 0, sizeof(num2));
        version1[len1] = '.'; len1 ++; version1[len1] = '\0';
        version2[len2] = '.'; len2 ++; version2[len2] = '\0';
        for(int i = 0; i < len1; i ++) {
            if(version1[i] != '.')
            for(int j = i; j < len1; j ++) {
                if(version1[j] != '.') continue;
                else {
                    cnt1 ++;
                    for(int k = i; k < j; k ++)
                        num1[cnt1] = num1[cnt1] * 10 + (version1[k] - '0');

                    i = j;
                    break;
                }
            }
        }
        for(int i = 0; i < len2; i ++) {
            if(version2[i] != '.')
            for(int j = i; j < len2; j ++) {
                if(version2[j] != '.') continue;
                else {
                    cnt2 ++;
                    for(int k = i; k < j; k ++)
                        num2[cnt2] = num2[cnt2] * 10 + (version2[k] - '0');

                    i = j;
                    break;
                }
            }
        }
        int flag = 0;
        int minn = min(cnt1, cnt2);
        for(int i = 1; i <= minn; i ++) {
            if(num1[i] == num2[i]) continue;
            else if(num1[i] > num2[i]) return 1;
            else return -1;
        }
        
        
            if(cnt1 == cnt2) return 0;
            else if(cnt1 > cnt2) {
                for(int i = minn + 1; i <= cnt1; i ++) {
                    if(num1[i]) return 1;
                }
            } else {
                for(int i = minn + 1; i <= cnt2; i ++) {
                    if(num2[i]) return -1;
                }
            }
            return 0;
    }
};