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);
}
can you explain this ?
ReplyDeleteint count=0;
ReplyDeleteif(n<1) return 0;
if(n%10==7) count++;
return count +count7(n/10);
count just switches back to 0 each method call...unne unnecessary memory usage
DeleteHow can I do this with while loop?
ReplyDeletepublic int count7(int n) {
ReplyDeleteif(n == 0) {
return 0;
}
if(n % 10 == 7) {
return 1 + count7(n / 10);
} else {
return count7(n / 10);
}
}
static int countOf7s=0;
ReplyDeletepublic 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?
r
ReplyDeleteint count = 0;
ReplyDeletewhile (n != 0) {
if (n % 10 == 7) {count++;}
n /= 10;
}return count;
no its asking recursive that require base step which is all positive numbers and recursive method means it calls the same method instead the method.
Deletepublic int count7(int n) {
ReplyDeleteif(n<0)throw new IllegalArgumentException("require positive number");
if(n==0)return 0;
else {
return (n%10==7)?1+count7(n/10):count7(n/10);
}