Java > Recursion-1 > count7 (CodingBat Solution)

Problem:

Given a non-negative int n, return the count of the occurrences of 7 as a digit, so for example 717 yields 2. (no loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).

count7(717) → 2
count7(7) → 1
count7(123) → 0


Solution:

public int count7(int n) {
  if (n < 1) return 0;
  if (n % 10 == 7) return 1 + count7(n/10);
  else return count7(n/10);
}


8 comments :

  1. int count=0;
    if(n<1) return 0;
    if(n%10==7) count++;
    return count +count7(n/10);

    ReplyDelete
    Replies
    1. count just switches back to 0 each method call...unne unnecessary memory usage

      Delete
  2. How can I do this with while loop?

    ReplyDelete
  3. public int count7(int n) {
    if(n == 0) {
    return 0;
    }

    if(n % 10 == 7) {
    return 1 + count7(n / 10);
    } else {
    return count7(n / 10);
    }
    }

    ReplyDelete
  4. static int countOf7s=0;
    public static int getCountOf7s(int intVal) {
    while(intVal>6){
    if(intVal==7 || intVal%10==7) {
    countOf7s++;
    }
    intVal=intVal/10;
    }
    return countOf7s;
    }

    Would this be bad code?

    ReplyDelete
  5. int count = 0;
    while (n != 0) {
    if (n % 10 == 7) {count++;}
    n /= 10;
    }return count;

    ReplyDelete

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