需要用数组中右侧元素的最大值替换该元素

Question

I need to replace every element in an array with the max element that belong to the right side set of elements in that array. I got a solution with the below code:

let arr1 = [5,0,5,7,9,4,8];
var temp = [];

for (var i = 1; i < arr1.length + 1; i++)
{
    if (i !== arr1.length)
    {
        temp = [...arr1].slice(i);
        arr1[i-1] = Math.max(...temp);
    }
    else
    {
        arr1[i-1] = arr1[arr1.length - 1];
    } 
}

console.log(arr1);

Is there any other better solution for solve this? You can see expected output on next sample:

Input : [5,0,5,7,9,4,8]
Output: [9,9,9,9,8,8,8]

Answer 1

You can use Array.map() . Note this approach, also, do not mutates your original array.

 let arr1 = [5,0,5,7,9,4,8]; let res = arr1.map( (n, idx, arr) => (idx < arr.length-1) ? Math.max(...arr.slice(idx+1)) : n ); console.log(res); 

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

Another alternative with O(n) performance can be approached using Array.reduceRight

 let arr1 = [5,0,5,7,9,4,8]; let max = arr1[arr1.length - 1]; let res = arr1.reduceRight( (acc, n, idx) => (acc[idx] = max, max = Math.max(n, max), acc), [] ); console.log(res); 

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

Answer 2

You can improve performance to O(n) (since your current implementation is O(n^2) ) by iterating backwards and keeping track of the max:

 var arr1 = [5, 0, 5, 7, 9, 4, 8]; var max = arr1[arr1.length - 1]; for (var i = arr1.length - 1; i >= 0; i--) { var curr = arr1[i]; arr1[i] = max; if (curr > max) max = curr; } console.log(arr1);