算法题 寻找 MountainArray 的 peek 求解惑

其实这个题是 https://leetcode-cn.com/problems/find-in-mountain-array/ 的一部分： 比如，array = [1,2,3,4,5,3,1]，这个数组是由严格递增和严格递减的两个有序数组构成的，它的 peek 就是 5. 现在我改一下条件，要么这个数组是严格递增和严格递减的两个有序数组构成的（这种情况返回 peek 的索引），要么这个数组就只是一个严格递增或严格递减的数组（返回-1 ）。

很明显，寻找 peek 可以用二分查找。

然后我发现这个算法有两种写法：

    public int findArrayPeek(int[] array) {
        int low = 0, high = array.length-1;

        int middle = -1;
        while (low <= high && (middle = (low + high)/2)-1 > 0 && middle+1 < array.length) {
            if (array[middle-1] < array[middle] && array[middle] > array[middle+1]) {
                return middle;
            } else if (array[middle-1] < array[middle] && array[middle] < array[middle+1]) {
                low = middle+1;
            } else {
                high = middle-1;
            }
        }

        return -1;
    }

    @Test
    public void findArrayPeekTest() {
        int[] array = {1,2,3,4,5,3,1};
        System.out.println(findArrayPeek(array));
        int[] array1 = {0,1,2,4,2,1};
        System.out.println(findArrayPeek(array1));
        int[] array2 = {0,1,2,4,5,1};
        System.out.println(findArrayPeek(array2));
        int[] array3 = {0,1,2,4,5,6};
        System.out.println(findArrayPeek(array3));
        int[] array4 = {6,5,4,3,2,1};
        System.out.println(findArrayPeek(array4));
    }

首先是标准的二分查找，但循环条件中必须要上索引的保护。

    public int findArrayPeek1(int[] array) {
        int low = 0, high = array.length-1;
        
        while (low < high-1) {
            int middle = (low + high)/2;
            if (array[middle-1] < array[middle] && array[middle] > array[middle+1]) {
                return middle;
            } else if (array[middle-1] < array[middle] && array[middle] < array[middle+1]) {
                low = middle;
            } else {
                high = middle;
            }
        }

        return -1;
    }

    @Test
    public void findArrayPeek1Test() {
        int[] array = {1,2,3,4,5,3,1};
        System.out.println(findArrayPeek1(array));
        int[] array1 = {0,1,2,4,2,1};
        System.out.println(findArrayPeek1(array1));
        int[] array2 = {0,1,2,4,5,1};
        System.out.println(findArrayPeek1(array2));
        int[] array3 = {0,1,2,4,5,6};
        System.out.println(findArrayPeek1(array3));
        int[] array4 = {6,5,4,3,2,1};
        System.out.println(findArrayPeek1(array4));
    }

然后是另外一种写法如上，它的处理过程从low = middle+1;变成了low = middle;，这里循环条件也必须得变，但这里不用加索引保护。

这两个算法，变化的只有对 low high 的处理，和循环条件。但感觉原理有点不一样，但我又说不清楚。。。（这种说不清楚的感觉，应该是我没有彻底弄懂~）

或者说，对 low high 的处理怎么影响这个过程的，相应的循环条件应该怎么写？