JavaScript 算法问题：如何使用递归来解决这样的组合问题？

Question

we have a digit letter map that looks like this

const digitsLetters = new Map([
    ["2", ['a', 'b', 'c']],
    ["3", ['d', 'e', 'f']],
    ["4", ['g', 'h', 'i']],
    ["5", ['j', 'k', 'l']],
    ["6", ['m', 'n', 'o']],
    ["7", ['p', 'q', 'r', 's']],
    ["8", ['t', 'u', 'v']],
    ["9", ['w', 'x', 'y', 'z']],
  ]);

The question asks us to return the all possible letter combinations that the number could represent. For example, if we have "23" then first digit 2 maps to ['a', 'b', 'c'] and second digit maps to ['d', 'e', 'f'] and we end up getting ["ad","ae","af","bd","be","bf","cd","ce","cf"] .

I have found a way to produce such a combination between two arrays.

// this will output ["ad","ae","af","bd","be","bf","cd","ce","cf"]
['a', 'b', 'c'].map(char1 => ['d', 'e', 'f'].map(char2 => char1 + char2)).flat(2)

So I thought I could just recursively apply this algorithm for such digit until I hit the last one. I think it is doable. However I had a hard time implementing the solution. Can someone please help me?

Answer 1

You can basically just use map method to get array from you Map data based on string param and them implement Cartesian product algo.

 const data = new Map([ ["2", ['a', 'b', 'c']], ["3", ['d', 'e', 'f']], ["4", ['g', 'h', 'i']], ["5", ['j', 'k', 'l']], ["6", ['m', 'n', 'o']], ["7", ['p', 'q', 'r', 's']], ["8", ['t', 'u', 'v']], ["9", ['w', 'x', 'y', 'z']], ]); function f(str, map) { const result = []; const arr = str.split('').map(c => map.get(c)) function r(data, n = 0, prev = []) { if (n === data.length) { return result.push(prev.slice()) } for (let i = 0; i < data[n].length; i++) { prev[n] = data[n][i] r(data, n + 1, prev.slice()) } } r(arr) return result; } console.log(f('23', data)) console.log(f('358', data))

Answer 2

Something like this

 const digitsLetters = new Map([ ["2", ['a', 'b', 'c']], ["3", ['d', 'e', 'f']], ["4", ['g', 'h', 'i']], ["5", ['j', 'k', 'l']], ["6", ['m', 'n', 'o']], ["7", ['p', 'q', 'r', 's']], ["8", ['t', 'u', 'v']], ["9", ['w', 'x', 'y', 'z']], ]); const foo = (arr, result = []) => { if (arr.length === 0) return result; const value = arr.shift(); if (result.length === 0) return foo(arr, value); const newResult = []; result.forEach((el) => { value.forEach((el2) => { newResult.push(el + el2); }); }); return foo(arr, newResult); }; const boo = (str) => foo(str.split('').map((symbol) => (digitsLetters.get(symbol)))); console.log(boo('')); console.log(boo('2')); console.log(boo('23')); console.log(boo('232'));

Answer 3

Let's consider the cartesian product f(E, F) with an example

assume E is like ['12', '13']

then you get candidates a and b , which I name as F = ['a', 'b']

f(E, F) = E x F would give ['12a', '13a', '12b', '13b']

(note that ExF is the same type of E , an array of string)

now you can recurse on the digits given

g([digit, ...otherDigits], E) => {
  candidates = m.get(digit)
  EF = cartesianProduct(E, candidates)
  return g(otherDigits, EF)
}

Note that the initialization on E should not be the empty array, but an array of length 1 whose sole element is the "neutral" element (empty string for strings)

 const data = new Map([ ["2", ['a', 'b', 'c']], ["3", ['d', 'e', 'f']], ["4", ['g', 'h', 'i']], ["5", ['j', 'k', 'l']], ["6", ['m', 'n', 'o']], ["7", ['p', 'q', 'r', 's']], ["8", ['t', 'u', 'v']], ["9", ['w', 'x', 'y', 'z']], ]); const f = (E, F) => E.flatMap(e => F.map(f => e + f)) const g = ([digit, ...otherDigits], E=['']) => digit? g(otherDigits, f(E, data.get(digit))): E console.log(g('23'.split(''))) console.log(g('234'.split('')))