Sword refers to the offer sixth "minimum number of rotating arrays" C + + implementation

Rotate array Minimum number of participants: 3258 time limit: 1 seconds space limit: 32768K pass ratio: 14.1% best record: 0 ms|0k (from Wocao) Topic Description

Moving a number of the first elements of an array to the end of the array, we call it the rotation of the array. Enter a rotation for a non-descending sequence, outputting the smallest element of the rotated array. For example, the array {3,4,5,1,2} is a rotation of {1,2,3,4,5}, and the minimum value of the array is 1.

This is also an orderly series, a look should be and two points, the fact is the case.

The topic gives a non-zero sequence, which means there may be duplicate numbers, to be noted.

To briefly analyze the possibility of arrays.

Rotated, not rotated, all elements are equal.

Look at the problem. The rotated array is divided into two halves, {3,4,5,} {1,2}.

You can see that if you rotate through the previous array, all elements must be greater than all of the following elements.

If not rotated, the first element must be smaller than the tail element.

If the middle element is greater than the tail element, the minimum value must be in the second half.

If the intermediate element is less than the first element, the minimum value must be in the first half.

Finally, there must be two elements left, followed by the minimum value, if the front is the minimum value is not rotating.

int Minnumberinrotatearray (vector<int> rotatearray) {//should not be free, because it is a problem in the cattle net, there is a set of test data is empty if (Rotatearray.empty ())
    {return 0;
    int left=0;
    int Right=rotatearray.size ()-1;
    if (Rotatearray[left]<rotatearray[right]) {return rotatearray[left];
        while (right>left) {int mid= (right+left)/2;
        int Middle=rotatearray[mid];
        int Leftnumber=rotatearray[left];
        int Rightnumber=rotatearray[right];
        if ((right-left) ==1) {return rightnumber;
            The values in the///left and right are all equal and cannot be determined either before or after the first one is a smaller than the previous one is the minimum if (middle==leftnumber&&middle==rightnumber) { for (int i=left;i<right;i++) {if (rotatearray[i]>rotatearray[i+1]) {return rota
                TEARRAY[I+1];
        }//All equal words directly return the leftmost value on the line returns Leftnumber;
        }else if (middle<leftnumber) {right=mid;
      }else if (middle>rightnumber) {      Left=mid; }
    }
}