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
s思路： 1. 由于version長度不規則，所以需要用while取數據。 取版本號時，由于版本可能超過int的范圍，需要不轉還直接比較嗎？ 2. bug:在寫的時候，剛開始判斷兩個string只要有一個遍歷完就退出循環，即：while(i
class Solution {public: int compareVersion(string version1, string version2) { // version1.push_back('.'); version2.push_back('.'); int i=0,j=0,m1=version1.size(),m2=version2.size(); //while(i<m1&&j<m2){//不專業的寫法 while(i<m1||j<m2){ int num1=0; if(i<m1&&version1[i]!='.'){ while(i<m1&&version1[i]!='.'){//加保護 num1=num1*10+version1[i]-'0'; i++; } }else i++; int num2=0; if(j<m2&&version2[j]!='.'){ while(j<m2&&version2[j]!='.'){//加保護 num2=num2*10+version2[j]-'0'; j++; } }else j++; //cout<<num1<<endl<<num2<<endl; if(num1>num2) return 1; if(num1<num2) return -1; } return 0; }};新聞熱點
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