Binary search under Java programming

Algorithm: This method is suitable when the amount of data is large. When using binary search, the data must be in order and not repeated. Basic idea: Assume that the data is sorted in ascending order. For a given value x, start the comparison from the middle position of the sequence. If the current position value is equal to x, the search is successful; if x is less than the current position value, then in the first half of the sequence Search; if x is greater than the current position value, continue searching in the second half of the sequence until it is found.

Suppose there is an array { 12, 23, 34, 45, 56, 67, 77, 89, 90 }. Now it is required to use the binary method to find the specified value and return it at the index of the array. If it is not found, return -1. The code is as follows:

package cn.sunzn.dichotomy;

public class DichotomySearch {
   public static void main(String[] args) {
       int[] arr = new int[] { 12, 23, 34, 45, 56, 67, 77, 89, 90 };
       System.out.println(search(arr, 12));
       System.out.println(search(arr, 45));
       System.out.println(search(arr, 67));
       System.out.println(search(arr, 89));
       System.out.println(search(arr, 99));
   }

   public static int search(int[] arr, int key) {
       int start = 0;
       int end = arr.length - 1;
       while (start <= end) {
           int middle = (start + end) / 2;
           if (key < arr[middle]) {
               end = middle - 1;
           } else if (key > arr[middle]) {
               start = middle + 1;
           } else {
               return middle;
           }
       }
       return -1;
   }
}

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