5, the array rotation

Subject description:

The beginning of an array of several elements moved to the end of the array, the array we call rotation.

A non-decreasing input array sort of a rotation, output rotation smallest element array.

For example, an array {3,4,5,1,2} {1,2,3,4,5} is a rotation of the array to a minimum.

NOTE: All the elements are given in greater than 0, if the array size is 0, return 0.

 

Problem-solving ideas:

Dichotomy answer this question,

mid = low + (high - low)/2

We need to consider three cases:

(1)array[mid] > array[high]:

This situation is similar to array [3,4,5,6,0,1,2], in this case a certain minimum number of mid right.

low = mid + 1

(2)array[mid] == array[high]:

This situation is similar to array [1,0,1,1,1] or [1,1,1,0,1], then the minimum number of bad judgment on the left mid

Or right, then I had a a try,

high = high - 1

(3)array[mid] < array[high]:

This situation is similar to array [2,2,3,4,5,6,6], then the minimum number of array necessarily [mid] or left mid

side. Because the right is inevitably increasing.

high = mid

Note that there is a hole: if only the last two numbers range to be queried, then the mid numerals will be directed at the front

For example array = [4,6]

array[low] = 4 ;array[mid] = 4 ; array[high] = 6 ;

If high = mid - 1, will generate an error, and therefore high = mid

However, the case (1), low = mid + 1 error will not

 

public int minNumberInRotateArray(int [] array) {
        int low = 0 ; int high = array.length - 1;   
        while(low < high){
            int mid = low + (high - low) / 2;        
            if(array[mid] > array[high]){
                low = mid + 1;
            }else if(array[mid] == array[high]){
                high = high - 1;
            }else{
                high = mid;
            }   
        }
        return array[low];
    }

 

 

Problem-solving ideas II:

  1, a first non-descending order array element to a minimum value, after the rotation, when the latter element is greater than the previous element, the element described later is minimum

  

int minNumberInRotateArray public (int [] Array) { 
     IF (Array == null || be array.length == 0) return 0; 
		
       for (int I = 0; I <-be array.length. 1; I ++) { 
		 IF (Array [ I]> array [I +. 1]) { 
			return array [I +. 1]; 
		 } 
	  } 
       return array [0]; // If not, rotation of the array does not occur, the minimum value of the first element 
}