如何检查数组是否在Java中增加数字序列？

Question

I am trying to check if an array has increasing sequences of numbers from 1 to n using a method in java and want to solve the problem using array. I have this code

public static int isIncreasingSequence(int[] a) {
        int start = 1, startIndex = 0, endIndex = 0;

    if(a[0] != 1)   return 0;

    while (start < a.length) {
        for (int j = startIndex; j < endIndex; j++) {
            if(a[j] != j)   return 0;
        }

        startIndex += start;
        endIndex += startIndex;
        start++;
    }

    return 1;
}

The method should return 1 if the array is in the sequence 1, 1, 2, 1, 2, 3, 1, 2, 3, 4 and so on. I can't get the desired output with my test array. Can anyone help me with it?

Answer 1

You were close, you just have to be careful about where you place your increments.

Naively, you need to increment to sub-sequence each time while keeping track of the absolute index in the array. Something like:

How does end keep track of the current sequence?

Because we iterate from 1 to end over and over again, and increment it by 1 until we reach the end of the array. We are also incrementing the index of the array while we do this.

At first, end = 1. So, we iterate from 1 to 1, and check each index in the array making sure a[0] == 1.

Next, end = 2. So, we iterate from 1 to 2, and check each index in the array making sure a[1] == 1, a[2] == 2.

Next, end = 3. So, we iterate from 1 to 3, and check each index in the array making sure a[3] == 1, a[4] == 2, a[5] == 3.

And so on.

public static int isIncreasingSequence(int[] a) {

        int end = 1; // Track the end of the current sequence
        int index = 0; // Track the index of the array

        while (index < a.length) {

            // Iterate througha larger and larger subsequence
            for (int i = 1; i <= end; i++) {
                if (a[index] != i) return 0;

                index++;
                if (index == a.length) break;
            }

            end++;
        }

        return 1;
    }