When the array is about to overflow, add 1/2 of the array, and then compare it with the largest constant of the array. If it exceeds the maximum size of the array, apply for a new oneInteger.MAX_VALUE的数组，然后把之前旧数组复制过来。
其中最难理解的是>>1, which is actually equivalent to dividing by 2.

Write a simple example and you will know after debugging it

public static void main(String[] args) {
    int oldCapacity = Integer.MAX_VALUE - 16;
    System.out.println(oldCapacity);
    int minCapacity = Integer.MAX_VALUE - 15;
    int maxSize = Integer.MAX_VALUE - 8;
    int newCapacity = oldCapacity + (oldCapacity >> 1);

    if (newCapacity - minCapacity < 0)
        newCapacity = minCapacity;
    if (newCapacity - maxSize > 0)
        newCapacity = hugeCapacity(minCapacity);
    // minCapacity is usually close to size, so this is a win:
    System.out.println(newCapacity);
}

private static int hugeCapacity(int minCapacity) {
    if (minCapacity < 0) // overflow
        throw new OutOfMemoryError();
    return (minCapacity > Integer.MAX_VALUE - 8) ?
            Integer.MAX_VALUE :
            Integer.MAX_VALUE - 8;
}

int newCapacity = oldCapacity + (oldCapacity >> 1);这句执行后如果超过int的最大值那么newCapacity will be a negative number. This requires understanding the principle of addition and subtraction of binary numbers.

The following four sentences are for handling newCapacitywhen overflow becomes a negative number

if (newCapacity - minCapacity < 0)
    newCapacity = minCapacity;
if (newCapacity - maxSize > 0)
    newCapacity = hugeCapacity(minCapacity);

I think you are talking about operating the same arraylist in multiple threads. At this time, the length overflow problem happened to occur. The problem is that arraylist is not thread-safe. The correct way is to use CopyOnWriteArrayList in java.util.concurrent