159. find minimum in rotated sorted array

public class  {
    
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        // write your code here
        
        Q: Does duplicate exist in the array?
        A: assume no duplicate exists.
        */
        
        1. 不能和nums【0】比
        如果nums[mid] < nums[0], 最小在mid左边(include)
        ?如果nums[mid] > nums[0], 最小在mid右边或者左边(无法判断)

        2. 要和nums【length-1】比
        nums[mid] != nums[length - 1]
        if nums[mid] < nums[length -1], target on the left of mid
        if nums[mid] > nums[length - 1], target on the right of mid
        */

        if (nums == null || nums.length == 0) {
            return -1;
        }

        int start = 0, end = nums.length - 1;
        int target = nums[nums.length - 1];
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] < target) {
                end = mid;
            } else if (nums[mid] > target) {
                start = mid;
            }
        }
        if (nums[start] <= nums[end]) {
            return nums[start];
        } else {
            return nums[end];
        }


    }
}