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Given a non-negative int n, return the sum of its digits recursively (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). sumDigits(126) → 9 sumDigits(49) → 13 sumDigits(12) → 3 ...Save, Compile, Run (ctrl-enter) |
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Difficulty: 318.0
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