I'm attempting to create a function that identifies the largest number in an array, but for some reason, I'm only able to retrieve the first number in the array.
function findLargest(numbers) {
var bigNum = 1;
for(i = 0; i < numbers.length; i++) {
if(numbers[i] > bigNum) {
bigNum = numbers[i];
}
return bigNum;
}
}
var numbers = [3, 4, 2, 6, 45, 775, 83, 5, 7];
findLargest(numbers);
You can utilize the Math.max function along with the spread operator to find the largest number in an array.
function findLargestNumber(arr) {
return Math.max(...arr);
}
var numbers = [3, 4, 2, 6, 45, 775, 83, 5, 7];
findLargestNumber(numbers);
console.log(findLargestNumber(numbers));The main issue was that the function was returning too early.
function findLargestNumber(numbers) {
// set the default result to null
var result = null;
if (numbers.length) {
// set the first number in the array as the default result
result = numbers[0];
for (i = 0; i < numbers.length; i++) {
if (numbers[i] > result) {
result = numbers[i];
}
} // <= the return statement was placed too early, outside the loop
}
return result;
}
var numbers = [3, 4, 2, 6, 45, 775, 83, 5, 7];
findLargestNumber(numbers);Check out this unique code snippet that helps find the largest number in an array:
function *largestNum(array) {
let result = -Infinity;
for (let value of array) yield result = value > result ? value : result;
}
Here's a simple function to get the maximum value of an array by selecting the last element:
function findMax(array) { return array[array.length - 1]; }
const numList = [8, 12, 4, 10, 67, 323, 42, 9, 15];
console.log(findMax([...largestNum(numList)]));
Remember to use [...] to convert the generator results into an array; alternatively, you can use Array.from.
function findLargest(numbers) {
var maximum = 1;
for (i = 0; i < numbers.length; i++) {
if (numbers[i] > maximum) {
maximum = numbers[i];
}
}
return maximum; // after going through the entire list of numbers!
}
var numbers = [3, 4, 2, 6, 45, 775, 83, 5, 7];
console.log(findLargest(numbers))
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