在数组中查找链接的递增值

Question

I'm trying to find all values in an array that would form a chain of incremented values - all referencing back to a certain starting value. Increments can go both "up" and "down".

array = [10, 2, 3, 5, 9, 11]

Starting with the number 2 should return:

[2, 3]

Starting with the number 10 should return:

[9, 10, 11]

There are of course plenty of inefficient ways of doing this, but I'm asking this here because doing this efficiently is important for my case and I'm such a JS newbie.

Answer 1

You can use Array.prototype.includes() to check if a number exists in an array. If the number is before the base reference add it using unshift , if after add it using push :

 var array = [10, 2, 3, 5, 9, 11]; function findChain(array, num) { if(!array.includes(num)) { return []; } const result = [num]; let before = num - 1; let after = num + 1; while(array.includes(before)) { result.unshift(before--); } while(array.includes(after)) { result.push(after++); } return result; } console.log('Ref 2 -', findChain(array, 2)); console.log('Ref 5 -', findChain(array, 5)); console.log('Ref 10 -', findChain(array, 10)); console.log('Ref 20 -', findChain(array, 20)); 

Answer 2

A quick solution :

var array = [10, 2, 3, 5, 9, 11, 14, 89, 12, 8];

var trouver = nombre => {
  var result = [];
  if (array.indexOf(nombre) !== -1) result.push(nombre);
  else return result;
  for(var chiffre = nombre+1; array.indexOf(chiffre) !== -1; chiffre++) result.push(chiffre);
  for(var chiffre = nombre-1; array.indexOf(chiffre) !== -1; chiffre--) result.push(chiffre);
  return result.sort((a,b) => a-b);
}

console.log(trouver(9)); //[ 8, 9, 10, 11, 12 ]

Answer 3

Another solution could be a double chained list for it.

 function getValues(array, value) { var object = Object.create(null), result, o; array.forEach(function (a) { object[a] = object[a] || { value: a, pre: object[a - 1] || null, succ: object[a + 1] || null }; if (object[a - 1]) { object[a - 1].succ = object[a]; } if (object[a + 1]) { object[a + 1].pre = object[a]; } }); o = object[value]; if (o) { result = []; while (o.pre) { o = o.pre; } while (o.succ) { result.push(o.value); o = o.succ; } result.push(o.value); } return result; } var array = [10, 2, 3, 5, 9, 11, 14, 89, 12, 8]; console.log(getValues(array, 2)); console.log(getValues(array, 10)); console.log(getValues(array, 42)); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

Answer 4

Try this approach

1. Sort the array first
2. Iterate array items one by one, keep pushing the start-counter if current-item value is not bigger than last-item by 1 , else reset the start-counter to current-index .

For Example :

  var arr = [10, 2, 3, 5, 9, 11]; function getAllSequences(arr) { arr.sort(function(a, b) { return a - b }); var startIndex = 0; var endIndex = 0; var lastItem = 0; var chains = []; arr.forEach(function(item, index) { if (index > 0) { if ((item - lastItem) > 1) { extractChain(chains, arr, startIndex, endIndex); startIndex = index; } else { endIndex = index; if (index == arr.length - 1) { extractChain(chains, arr, startIndex, endIndex); } } } lastItem = item; }); return chains; } console.log(getAllSequences(arr)); function extractChain(chains, arr, startIndex, endIndex) { var value = arr.slice(startIndex, endIndex + 1); if (value.length > 0) { chains.push(value); } }