Binary find the deformation of the algorithm; The front and back end are orderly, the whole disorder

The ' Package Mainimport ("FMT")//is divided into 2, certainly half is ordered by the Func binarysearchdef (data []int, key int, low int, high int) int {if Len (d ATA) = = 0 {return-1}//Note the critical value for low <= high {mid: = low + (high-low)/2fmt. Println (mid) fmt. Println (Data[mid]) if key = = Data[mid] {return mid}//if key is smaller than intermediate value if key < Data[mid] {if key = = Data[low] {return low}// If key is smaller than the median value, but is greater than the value in the low position if key > Data[low] {//If key is smaller than the median value, the value above the low position is larger, and the element between lower and mid is unordered if mid-1 >= low &amp ;& Data[low] >= data[mid-1] {return binarysearchdef (data, key, Low, mid-1)}//orderly reverse the other half high = Mid-1}if key < da Ta[low] {if mid-1 >= low && Data[low] >= data[mid-1] {return binarysearchdef (data, key, Low, mid-1)}low = mi D + 1}}if key > Data[mid] {if key = = Data[high] {return high}if key > Data[high] {if mid+1 <= high && dat A[MID+1] >= Data[high] {return binarysearchdef (data, key, mid+1, high)}//in order to reverse the other half, because there must be no corresponding value in the section high = Mid-1}if key < Data[high] {if mid+1 <= high && data[mid+1] >= Data[high] {return binarysearchdef (data, key, mid+1, high)}low = mid + 1}}}return-1}func Main () {data: = [] INT{100, 102, 104, 106, 108, 2, 118, 121, 122, 1,, 3, 4, 5, 6, 7, 8, 9, 10}i: = Binarysearchdef (da TA, 3, 0, Len (data)-1) fmt. Println (i)} "90 reads