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) → 1 count8(818) → 2 count8(8818) → 4 ...Save, Compile, Run public int count8(int n) { } See Also: Example Code Java help: If Boolean Logic | Strings | While and For Loops | Arrays and Loops prev  |  next  |  chance   |  JavaBat  >  Recursion-1

Forget It! -- delete my code for this problem 324.0

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