题目：把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如，数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。
思路：分为两种情况，如果第一个元素和最后一个元素不相等，则用二分法；如果第一个元素等于最后一个元素，则用顺序查找。
解决方案：

public class Question11 {
    public static int min(int[] numbers, int length){
        if (numbers == null || length < 0) try {
            throw new Exception("Invalid parameters");
        } catch (Exception e) {
            e.printStackTrace();
        }

        int index1 = 0;
        int index2 = length - 1;
        int indexMid = index1;
        while (numbers[index1] >= numbers[index2]){
            if (index2-index1 == 1){
                indexMid = index2;
                break;
            }
            indexMid = (index1 + index2) / 2;
            if (numbers[index1] == numbers[index2] && numbers[indexMid] == numbers[index1]){
                return minInOrder(numbers, index1, index2);
            }

            if (numbers[indexMid] >= numbers[index1]){
                index1 = indexMid;
            }

            if (numbers[indexMid] <= numbers[index2]){
                index2 = indexMid;
            }
        }
        return numbers[indexMid];
    }
    private static int minInOrder(int[] numbsers, int index1, int index2){
        int result = numbsers[index1];
        for (int i=index1 + 1; i<=index2; i++){
            if (result > numbsers[i]){
                result = numbsers[i];
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] numbers = new int[]{3,4,5,1,2};
        int[] numbers1 = new int[]{1,0,1,1,1,1};
        System.out.println(min(numbers, numbers.length));
        System.out.println(min(numbers1, numbers1.length));
    }
}