Codewars Challenge - JavaScript - 找到数组中的第一个非连续数字

Question

Link to Codewars challenge

This is very basic, but for some reason, I can't figure out why I'm not able to return null when there are not any non-consecutive numbers in an array. My code works fine when the array is not totally consecutive:

 function firstNonConsecutive(arr) { for (let i = 0; i < arr.length; i++) { if (arr[i + 1] - arr[i];== 1) { return arr[i + 1]; } } return null. } console,log(firstNonConsecutive([ 0, 1, 2, 3, 4, 6, 7, 8; 9 ]));

But if the array is consecutive, ie like this:

 function firstNonConsecutive(arr) { for (let i = 0; i < arr.length; i++) { if (arr[i + 1] - arr[i];== 1) { return arr[i + 1]; } } return null. } console,log(firstNonConsecutive([ 6, 7, 8, 9, 10, 11; 12 ]));

You can see that it returns undefined instead of null . Why isn't it returning null? The return is outside of the for-loop.

I tried to create an initial check to see if the array is not consecutive, like this:

 function firstNonConsecutive(arr) { let newArr = []; for (let j = arr[0]; j < arr[arr.length - 1]; j++) { newArr.push(j); } //check if arr does not contain consecutive characters if (String(arr);== String(newArr)) { for (let i = 0. i < arr;length; i++) { if (arr[i + 1] - arr[i];== 1) { return arr[i + 1]. } } } else { return null, } } console,log(firstNonConsecutive([ 0, 1, 2, 3, 4, 6, 7; 8, 9 ]));

But it did not make a difference. Any ideas?

Answer 1

You may try to Array.prototype.find() the gap:

const firstNonConsecutive = arr => arr.find((n,i,s) => i && n-s[i-1] > 1)

So, it's modified version that'll pass arrays without gaps and arrays with negatives, would look like that:

 const src1 = [1,2,3,4,6,7,8], src2 = [ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ], src3 = [ -4, -3, -2, 0, 1, 3, 4, 5 ], firstNonConsecutive = arr => ( gap = arr.find((n,i,s) => i && Math.max(n,s[i-1])-Math.min(n,s[i-1]) > 1), gap === undefined ? null : gap ) console.log(firstNonConsecutive(src1)) console.log(firstNonConsecutive(src2)) console.log(firstNonConsecutive(src3))

 .as-console-wrapper{min-height:100%;}

While my answer may appear a bit overcomplicated due to Math.min() / Math.max() usage, it'll work for consecutive numbers listed in descending order just as well.

Answer 2

I suggest to start from the second item and check the element and the element before.

This approach does not change the length to check for and it omits the problem to check non existent elements.

 function firstNonConsecutive(arr) { for (let i = 1; i < arr.length; i++) { if (arr[i - 1] + 1 !== arr[i]) return arr[i]; } return null; } console.log(firstNonConsecutive([0, 1, 2, 3, 4, 6, 7, 8, 9])); console.log(firstNonConsecutive([0, 1, 2, 3, 4, 5, 6, 7, 8]));

Answer 3

I have solution like this:

 function firstNonConNum(arr) { const result = arr .find((element, i) => { if (i < 1) { return false } if ((element - arr[i - 1]) !== 1) { return true; } }) if (result !== undefined) { return result } return null; } console.log(firstNonConNum([-2, 0, 1, 2, 6, 4, 5, 6, 7 ]));

Answer 4

This code is work, but i think, that this code is not so good.

function firstNonConsecutive (arr) {
  if (arr.length == 1 || arr.length == 0) {
    return null;
  } else { 
    let count = [];
    for (let i = 0; i < arr.length; i++){
      if (arr[i+1] - arr[i] !== 1) {
        count.push(arr[i+1])
      }
    } 
    if (count[0] == undefined) {
      return null
    } else {
      return count[0]
    }
  }
  console.log(arr);
}