Compare Version Numbers--LeetCode

標籤：c++   leetcode   演算法   

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

思路：每次比較小數點之前的數，看能不能分出伯仲。就是比較“版本號碼”大小，給定的版本號碼version1和version2是字串類型的，當version1>version2的時候，返回1，反之返回-1。

這道題屬於細節處理題，除了字串處理繁瑣一點之外沒有什麼。

解題思路：先分別將version1、version2字串按‘.‘分割成多個子串，每個子串轉化成整型存入容器。最後再比較兩個容器中對應位置的數的大小。當然需要考慮它們長度不同的情況。

注意點：

（1）version的形式：10.23.1、01.23、1.2.3.4....諸如

int compareVersion(string& s1, string& s2) {    int i=0,j=0;    int temp1,temp2;    while(i<s1.length() && j<s2.length())    {       temp1=0;       temp2=0;       for(;i<s1.length();i++)         if(s1[i]!='.')         temp1  = temp1*10 + (s1[i]-'0');       for(;j<s2.length();j++)         if(s2[j]!='.')         temp2  = temp2*10 + (s2[j]-'0');       if(temp1> temp2)        return 1;       else if(temp1<temp2)        return -1;       else       {           i++;           j++;       }                      }        if(i>=s1.length() && j< s2.length())      return -1;    if(i<s1.length() && j>=s2.length())      return 1;    if(i>=s1.length() && j>=s2.length())      return 0;}

Compare Version Numbers--LeetCode