Sword finger offer | minimum number of rotation array

There is a length of n Non descending array of , such as [1,2,3,4,5], Rotate it , That is, move the first elements of an array to the end of the array , Into a rotating array , For example, it became [3,4,5,1,2], perhaps [4,5,1,2,3] In this way . Excuse me, , Given such a rotation array , Find the minimum value in the array

Ideas ：

Although after rotation , But the first and second paragraphs are orderly , So consider using dichotomy to solve .
send middle = （start+end）/2, take middle and end Compare ：
If Greater than , The result is on the left ; If Less than , The result is on the right ;
If be equal to , It is impossible to judge which side （ for example 10111 11101 Both sides are possible ） Need to shrink end Keep looking for .
Why not use the left start and middle Compare ？
for example 12345 3>1 , The minimum value is on the left , 34512 5>3 But the minimum value is on the right . It's impossible to judge .

Code ：

public int minNumberInRotateArray(int [] array) {
    
  if(array.length == 0){
    
   		return 0;
   }
   int start = 0;
   int end = array.length-1;

    while(start < end){
    
        int middle = (start+end)>>>1;
        if(array[middle] < array[end]){
    
            end = middle;
        }else if(array[middle] > array[end]) {
    
           start = middle + 1;
        }else {
    
            end --;
        }
    }
    return array[end];
}

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Living without an aim is like sailing without a compass.

—— J. Ruskin