因此,我有这段代码可以提示两侧的空格分隔输入和三角形的斜边。如果三角形正确,则main下面的方法应返回true,否则返回false。

出于某种原因,即使我知道输入的测量值是直角三角形,它仍然打印出三角形不是直角三角形。

我已经尝试在这段代码中检测到逻辑错误一段时间了,可以帮忙吗?

import java.util.Scanner;


public class VerifyRightTriangle { public static void main(String[] args) { Scanner sc = new Scanner(System.in);

    System.out.print("Please enter the two sides and the hypotenuse: ");
    String input = sc.nextLine();

    String[] values = input.split(" ");

    int[] nums = new int[values.length];

    for (int i = 0; i < nums.length; i++) {
        nums[i] = Integer.parseInt(values[i]);
    }

    double s1 = nums[0];
    double s2 = nums[1];
    double hyp = nums[2];

    System.out.printf("Side 1: %.2f Side 2: %.2f Hypotenuse: %.2f\n", s1, s2, hyp);

    boolean result = isRightTriangle(s1, s2, hyp);

    if (result == true) {
        System.out.println("The given triangle is a right triangle.");
    } else if (result == false) {
        System.out.println("The given triangle is not a right triangle.");
    }
}

/**
 * Determine if the triangle is a right triangle or not, given the shorter side, the longer
 * side, and the hypotenuse (in that order), using the Pythagorean theorem.
 * @param s1 the shorter side of the triangle
 * @param s2 the longer side of the triangle
 * @param hyp the hypotenuse of the triangle
 * @return true if triangle is right, false if not
 */

private static boolean isRightTriangle(double s1, double s2, double hyp) {
    double leftSide = s1 * s1 + s2 * s2;
    double rightSide = hyp * hyp;
    if (Math.sqrt(leftSide) == Math.sqrt(rightSide)) {
        return true;
    } else {
        return false;
    }
}


}

最佳答案

浮点计算不精确,这就是为什么数字不匹配的原因,请尝试使用BigDecimal而不是double

编辑:
再次思考,因为您已经在解析整数
nums[i] = Integer.parseInt(values[i]);

只需使用int代替double,而不使用Math.sqrt,只需使用leftSide==rightside

关于java - Java-找不到逻辑错误,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/12238012/

10-10 05:01