Find Min In Rotated Sorted Array2，包含重复数字的反转序列找最小值。

public int findMin(int[] nums)
    {
        return findMin(nums, 0, nums.length - 1);
    }
    public int findMin(int[] nums, int left, int right)
    {
        int mid = (left + right)/2;
        if(left == right)
        {
            return nums[left];
        }
        if(nums[left] < nums[right])
        {
            return nums[left];
        }
        if((left - right) == 1)//这一步必须要有，因为会涉及到left+1，right-1，例如序列为11.
        {
            return Math.min(nums[left], nums[right]);
        }
        if(nums[left] == nums[mid] && nums[mid] == nums[right])
        {
            return findMin(nums, left + 1, right - 1);
        }
        else if(nums[left] <= nums[mid])
        {
            return findMin(nums, mid+1, right);
        }
        else
        {
            return findMin(nums, left, mid);
        }
    }
    //迭代方法
    public int findMin2(int[] nums, int left, int right)
    {
        while(left <= right)
        {
            int mid = (left + right)/2;
            if(left == right)
            {
                return nums[left];
            }
            if(nums[left] < nums[right])
            {
                return nums[left];
            }
            if(nums[right] == nums[mid] && nums[mid] == nums[right])
            {
                left ++;
                right --;
            }
            else if(nums[left] <= nums[mid])
            {
                left = mid + 1;
            }
            else
            {
                right = mid;
            }
        }
        return nums[left];
    }