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];
     }