id/email | |

password | |

forgot password | create account | |

Or use |

about | help | code help+videos | done | prefs |

Recursion-1 > count8

prev | next | chance

Forget It! -- delete my code for this problem### Java Help

### Misc Code Practice

prev | next | chance

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 |

Forget It! -- delete my code for this problem

Progress graphs:

Your progress graph for this problem

Random user progress graph for this problem

Random Epic Progress Graph

- Java Example Solution Code
- Java String Introduction (video)
- Java Substring v2 (video)
- Java String Equals and Loops
- Java String indexOf and Parsing
- Java If and Boolean Logic
- If Boolean Logic Example Solution Code 1 (video)
- If Boolean Logic Example Solution Code 2 (video)
- Java For and While Loops
- Java Arrays and Loops
- Java Map Introduction
- Java Map WordCount
- Java Functional Mapping
- Java Functional Filtering

- Code Badges
- Introduction to Mod (video)
- MakeBricks problem and solution (video x 2)
- FizzBuzz the famous code interview question (video)

Difficulty: 324.0

Copyright Nick Parlante 2016 - privacy