Binary Search 相关的case

Case 1: 在一个sorted array里面找出target value的 index，如果target value 出现了多次，返回任何一个就可以了。

Cas 2:  Find the smallest number which is greater than the target

 

Case 3: Find the largest number which is smaller than the target

Case 4: Find the left most index of the target value in the sorted array. If not found, return the index of the smallest number which is greater than the target.

    private static int leftMostIndex(int[] arr, int target) {
        int left = 0, right = arr.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

 Case 4: Find the right most index of the target value in the sorted array. If not found, return the index of the largest number which is smaller than the target.

    private static int rightMostIndex(int[] arr, int target) {
        int left = 0, right = arr.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] <= target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return right;
    }