比较 arrays 中的值，如果任何值匹配，则增加第一个数组中的值

Question

I've got this code to check two arrays for any matching values, and if any values do match I ++ everything in array1

I'm wondering if there is a better way to do this, since I feel like this is a lot of looping, and I'll eventually have 5 arrays i need to compare against each other

Any help would be greatly appreciated!

const array1 = [2, 9];

const array2 = [2, 5, 9];

function checkMatch(a, b) {
  for (let i = 0; i < a.length; i++) {
    for (let e = 0; e < b.length; e++) {
      if (a[i] === b[e]) a[i]++;
    }
  }
  return a;
}

console.log(checkMatch(array1, array2))

Answer 1

If you know that all arrays are sorted, then you can use the following approach -

CODE -

 const array1 = [2, 9]; const array2 = [2, 5, 9]; function checkMatch(a, b) { let i = 0, j = 0; while (i < a.length && j < b.length) { if (a[i] === b[j]) { a[i]++; j++; } else if (a[i] < b[j]) { i++; } else j++; } return a; } console.log(checkMatch(array1, array2))

Explanation -

The above approach will have a time complexity of O(N+M) while your could would have had a time complexity of O(N*M) .

In the above function, you are taking advantage of the fact that arrays are sorted. So when a[i] < b[j] , you know that you have to increment index i to get a value that can potentially be equal to or greater than b[j] . The same (but in the reverse way) is the case when a[i] > b[j] . So, this approach reduces the overall time-complexity of your code and would increases the overall efficiency.

Hope this helps!

Answer 2

You can use includes method to check if all elements of a list (a) exits in all other arrays you need to compare. Then you can update the value in list (a)

 function checkMatch(a, b){ for (let i = 0; i < a.length; i++) { if(b.includes(a[i])){ a[i]++; } } return a; }; const array1 = [2, 9]; const array2 = [2, 5, 9]; console.log(checkMatch(array1, array2))

Answer 3

You can simply make use of map here:

 var array1 = [2, 9]; var array2 = [2, 5, 9]; var result = array1.map(n=>(array2.includes(n)? n++: n, n)); console.log(result);

Answer 4

You could take a Set and map the first array with the value plus the check for this value of the set.

 function checkMatch(a, b) { var values = new Set(b); return a.map(v => v + values.has(v)); } console.log(checkMatch([2, 9], [2, 5, 9]));