旋转数组查找

程序员面试金典里的解法才是正解，逻辑上没有漏洞，其他书籍的解法是有问题的：

无论数组是否有相同数字。都可以使用下面解法：

def search(arr, left, right, x):
    if left > right: return -1
    mid = (left+right)/2
    if x == mid:
        return mid
    
    # 你先看下最外围的三个if， 逻辑完备，>, <, = 都考虑到了!!!
    if arr[left] < arr[mid]:
        # 再看里面的if，都是在确定究竟哪一段有序
        if x>= arr[left] and x<arr[mid]:
            return search(arr, left, mid-1, x)
        else:
            return search(arr, mid+1, right, x)
    elif arr[mid] < arr[left]:
        if x>arr[mid] and x<=arr[right]:
             return search(arr, mid+1, right, x)
        else:
             return search(arr, left, mid-1, x)
     else:
         if arr[mid] == arr[right]: # 说明arr[mid]==arr[left]==arr[right]，两边都有相等数据，只能两边继续二分
              result = search(arr, left, mid-1, x)
              if result == -1:
                  return search(arr, mid+1, right, x)
              else:
                  return result
         else: # 说明 arr[mid]!=arr[right]，仅left这边遇到相等
             return search(arr, mid+1, right, x) 

这段代码非常关键：

else:
        if arr[mid] == arr[right]: # 说明arr[mid]==arr[left]==arr[right]，两边都有相等数据，只能两边继续二分
             result = search(arr, left, mid-1, x)
             if result == -1:
                 return search(arr, mid+1, right, x)
             else:
                 return result
        else: # 说明 arr[mid]!=arr[right]，仅left这边遇到相等
            return search(arr, mid+1, right, x)

同理，对于求解旋转数组最小值的解法：

def get_min_of_rotation(arr, left, right):
    assert left <= right
    if left == right: return arr[left]
    mid = (left+right)/2
    if arr[mid] > arr[left]:
        return min(arr[left], get_min_of_rotation(arr, mid+1, right))        
    elif arr[mid] < arr[left]:
        return min(arr[mid], get_min_of_rotation(arr, left, mid-1))            
    else:
        if arr[mid] != arr[right]:
            return min(arr[mid], get_min_of_rotation(arr, mid+1, right))  
        else:
            return min(arr[mid], get_min_of_rotation(arr, mid+1, right), get_min_of_rotation(arr, left, mid-1)) 