Binary search binary search method

Search method to search while narrowing the range in half

merit

Search speed is fast

Demerit

Data needs to be sorted in advance

`BinarySerch.java`



class BinarySerch{
  public static void main(String[] args){
   //Sorted array to search
   int a[] = {1,3,6,8,10};
   //Value to search
   int serchValue = 3;
   //The sequence number of the found data The initial value is an error value (-1）
   int find = -1;

   //Leftmost and rightmost numbers to search
   int left = 0;
   int right = a.length-1;
   //Repeat when there is data on the left and right edges to check
   while(left <= right){
     //Specify the position to check the number in the middle of the left and right
     int middle = (int)((left + right)/2);
     //Compare the value of the position to look for with the position to look for
     if(a[middle] == serchValue){
        //If they are the same, save the number and end the iteration
        find = middle;
        break;
    }else if (a[middle] < serchValue){
    //If it is smaller than the value you are looking for, move the left edge
     left = middle + 1;
    }else {
    //If it is larger than the value you are looking for, move the right edge
    right = middle - 1;
    }
  }
  //Display search results
  System.out.println("search results:" + find);
 }
}

Reference book [Enjoyable learning algorithm and programming pictorial book](https://www.amazon.co.jp/%E6%A5%BD%E3%81%97%E3%81%8F%E5%AD%A6%E3%81] % B6-% E3% 82% A2% E3% 83% AB% E3% 82% B4% E3% 83% AA% E3% 82% BA% E3% 83% A0% E3% 81% A8% E3% 83% 97% E3% 83% AD% E3% 82% B0% E3% 83% A9% E3% 83% 9F% E3% 83% B3% E3% 82% B0% E3% 81% AE% E5% 9B% B3% E9% 91% 91-% E6% A3% AE-% E5% B7% A7% E5% B0% 9A-ebook / dp / B01N56TW48)