Practice›Java Fundamentals›Module 5. Arrays & Lists›5.2 Array algorithms

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5.2 Array algorithms — finding the maximum

Lesson

Tracking the largest so far

To find the biggest element, keep a variable for the largest seen so far. Start it at the first element (arr[0]), then compare each remaining element and update when you find something bigger. Starting at arr[0] (not 0) keeps it correct even for all-negative arrays.

Worked example

int[] nums = {3, 7, 2, 9, 4};
int max = nums[0];
for (int i = 1; i < nums.length; i++) {
    if (nums[i] > max) {
        max = nums[i];
    }
}
System.out.println(max);   // 9

The loop starts at index 1 because max already holds element 0.

▶ Try it: run this to find the largest value. Change the numbers and re-run.

int[] nums = {3, 7, 2, 9, 4};
int max = nums[0];
for (int i = 1; i < nums.length; i++) {
    if (nums[i] > max) {
        max = nums[i];
    }
}
System.out.println(max);

To find the maximum, what should max start at?

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What is the maximum of {3, 7, 2, 9, 4}?

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Inside the loop, when do you update max?

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Type the maximum this prints:

int[] a = {5, 2, 8, 1};
int max = a[0];
for (int i = 1; i < a.length; i++) {
    if (a[i] > max) { max = a[i]; }
}
System.out.println(max);

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True or false?

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