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Java > Recursion-1 > count8
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 Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. 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). count8(8) → 1count8(818) → 2count8(8818) → 4...Save, Compile, Runpublic int count8(int n) { } Show output only (no red/green) prev  |  next  |  chance   |  CodingBat  >  Recursion-1

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