Define dichotomy lookup

int[ ] nums={1,2,4,5,6,8,9,12,23,45,67,68,79,80,90,200};

public static int binarySearch(int[ ] nums,int v,int start,int end){

}

Ideas ：

1. Calculate the index of the middle position

2. Compare the searched value with the middle position , If it is larger than the search location , Recursively call （ Array , middle +1,end）;

If small ,（ Array ,start, In the middle -1）

3. Until the search value is equal to the middle position （ find ）, perhaps start Greater than or equal to end（ Not found ）, Recursion ends

Mode one ： 

public class BinarySearch {
    public static int binary(int[] nums,int start,int end,int values) {
        if(start>end) {
            return -1;
        }
        int mid=(start+end)/2;
        if(values<nums[mid]) {

            return  binary(nums, start, mid-1, values);
        }else if(nums[mid]<values) {
            return binary(nums, mid+1, end, values);

        }else {
            return mid;
        }
    }
    public static void  main(String[] args) {
        int [] nums= {1,2,4,5,6,8,9,12,23,45,67,68,79,80,90,200};
        System.out.println(" The index of this number is ："+binary(nums,0,nums.length-1,8));
    }
}

Mode two ： 

public static int binarySearch(int[] nums,int v,int start,int end) {
    //1. The condition for ending recursion 
    if(start>end) {
        return -1;
    }
    //2. Business logic 
    // Find the middle point ;
    int mid=(start+end)>>1;
	if(nums[mid]==v) {
		return mid; // find 
	}else if(nums[mid]>v) {//3. Recursively call down 
		return binarySearch(nums,v,start,mid-1);
	}else {
		return binarySearch(nums,v,mid+1,end);
	}
}