面试题8：旋转数组的最小数字

//面试题8：旋转数组的最小数字

//递增排序的{1,2,3,4,5}----->旋转为 {3,4,5,1,2}
//求最小的数字

package com.hb.jzoffer;

import offer.utilities.ArrayUtils;

//面试题8：旋转数组的最小数字

//{1,2,3,4,5}----->旋转为 {3,4,5,1,2}

publicclass MinNumberInRotatedArray_8 {

public  int  Min(int[] array ) throws Exception{
        int  length = array.length ;
        if(length < 0){
            Exception e = new Exception();
            throw e;
        }
        int  index1 = 0 ; 
        int  index2 = length - 1;
        int  indexMid = index1 ; //如果下面while循环里面的不成立，说明数组是有序的，没有旋转，所以indexMid初始化为Index1
        while(array[index1] >= array[index2]){
            if(index2 - index1 == 1){
                indexMid = index2;
                break ;
            }
            indexMid = (index1 + index2) / 2 ;
            
            //如果下标为index1 、index2 和 indexMid 指向的三个数字相等，
            //则只能用书序查找
            if(array[index1] == array[index2] && array[indexMid] == array[index1]){
                return MinInOrder(array , index1 , index2);                
            }
            
            if(array[indexMid] >= array[index1]){
                index1 = indexMid ;
            }else if(array[indexMid] <= array[index2]){
                index2 = indexMid ;
            }    
        }
        
        return array[indexMid];    
        
    }
    
    public int MinInOrder(int[] array , int index1 , int index2){
        int  result = array[index1];
        for(int i = index1 + 1 ; i <= index2 ; i++){
            if(result > array[i]){
                result = array[i] ;
            }            
        }        
        return result;
    }
    
    public static void main(String[] args) throws Exception {
        int[] array = {3,4,5,1,2};
        int[] array1 = {1,1,1,0,1};
        MinNumberInRotatedArray_8 findMid = new MinNumberInRotatedArray_8();
        int min = findMid.Min(array);
        int min1 = findMid.Min(array1);
        ArrayUtils.printArray(array);
        System.out.println(min);
        ArrayUtils.printArray(array1);
        System.out.println(min1);
        
    }
    
    

}