二进制搜索不会停止吗？

问题描述

我具有以下二进制搜索方法和以下驱动程序代码。我搜索的两个显示在输出中的值都存在于数组中。

搜索方法

    //Method for binary search. This method will also cut our array
    public static int binarySearch(int[] nums, int x) {
        //Bounds
        int l = 0, r = nums.length - 1;
        //While the size of the array is not 1
        while (l  nums[m]) l = m + 1;
            //Else, ignore right half
            else r = m - 1;
        }
        //If we didn't find the element
        return -1;
    }

驱动程序代码和输出

public class searcher {
    public static void main(String[] args) {
         /* Initialize a new scanner for user input, initialize random for the
        computer to pick a number */
        Scanner s = new Scanner(System.in);
        //Variable for user input
        int guess;
        //Do-while loop
        do {
            System.out.println("Enter a number to search for (0 to quit): ");
            //Get the user's guess
            guess = s.nextInt();
            //Search for the guess in the array of numbers
            int i = binarySearch(nums, guess);
            System.out.println(i);
            //If the number is not found
            if (i == -1) {
                System.out.println("Your number does not occur in this list.");
            }
            //If it is
            else {
                System.out.println("Your number occurs at position " + i);
            }
        } while (guess != 0);
    }
}
/*
Output
Enter a number to search for (0 to quit): 
1
1
Your number occurs at position 1
Enter a number to search for (0 to quit): 
90 
            

我期望找到找到的输入数字的索引，如果没有找到，该方法应该返回-1，这样我就可以打印出未找到的数字

思路一：

int m = l + (r - 1) / 2; // this is not correct, you are subtracting "1"

您需要减去“ left”（为清楚起见，对变量名进行了编辑）：

int mid = left + (right - left) / 2;

或者更好一点：

int mid = (left+ right) >>> 1;

思路二：

您减去1两次。

r = nums.length - 1;

然后

int m = l + (r - 1) / 2;

应该是

int m = l + (r - l) / 2;