2180. Count Integers With Even Digit Sum

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problem

solution

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class Solution {
public:
    bool isEvenDigit(int n){
        int digit = 0;
        while(n){
            digit+=(n%10);
            n/=10;
        }
        return digit%2==0;
    }
    int countEven(int num) {
        // option 1  brute force time complexity O(n^2)
        int count=0;
        for(int i=1;i<=num;++i){
            count+= isEvenDigit(i)?1:0;
        }
        return count;
        
    }
};

analysis

  • time complexity O(nlogn)
  • speed complexity O(1)