2224. Minimum Number of Operations to Convert Time

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problem

solution

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class Solution {
public:
    int convertTime(string current, string correct) {
        int diff_hour = 0, diff_min = 0;
        for(int i=0;i<2;++i){
            diff_hour = diff_hour*10 + (correct[i]-'0') - (current[i] - '0');
        }
        int count = diff_hour;
        for(int i=3;i<correct.size() ; ++i){
            diff_min = diff_min*10 + (correct[i]-'0') - (current[i] - '0');
        }
        if(diff_min<0){
            // 借位
            count--;
            diff_hour--;
            diff_min+=60;
            
        }
        while(diff_min>=15){
            diff_min-=15;
            count++;
        }
        while(diff_min>=5){
            diff_min-=5;
            count++;
        }
        while(diff_min>=1){
            diff_min-=1;
            count++;
        }
        return count;
        
    }
};

analysis

  • time complexity O(1)
  • space complexity O(1)