二分查找算法——Java版

这是将网上的C++版的改写成Java的

理论上把函数f()改一下就可以计算任意方程的0点。

仅供参考，可以将其写成递归的。

 

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class ErFenFa {

	static int m;
	static double x1, x2, x3;

	public static double f(double x) {
		double y;
		y = x * x * x - x - 1;
		return y;
	}

	public static void main(String[] args) throws NumberFormatException,
			IOException {
		double a, b;
		System.out.println("please input two number a,b");
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		a = Double.valueOf(br.readLine());
		b = Double.valueOf(br.readLine());

		double x1 = a;
		double x2 = b;
		double x3;

		for (int i = 1;; i++) {
			if (f(x1) == 0) {
				m = (int) (x1 * 100);
				x1 = m / 100;
				System.out.println("方程的根是x=" + x1 + "，迭代次数=" + i);
			}
			if (f(x2) == 0) {
				m = (int) (x2 * 100);
				x2 = m / 100;
				System.out.println("方程的根是x=" + x2 + "，迭代次数=" + i);
			}
			if (f(x1) * f(x2) > 0) {
				System.out.println("方程无解！" + "，迭代次数=" + i);
				return;
			} else if (f(x1) * f(x2) < 0) {
				x3 = (x1 + x2) / 2; // return 1;

				if (f(x3) == 0) {
					m = (int) (x3 * 100);
					x3 = m / 100;
					System.out.println("方程的根是x=" + x3 + "，迭代次数=" + i);
				}
				if (f(x3) * f(x1) < 0) {
					x2 = x3;
					if (Math.abs(x1 - x3) < 0.00001) {
						m = (int) (x2 * 100);
						x2 = m / 100;
						System.out.println("方程的根是x=" + x2 + "，迭代次数=" + i);
						break;
					}
				}
				if (f(x3) * f(x2) < 0) {
					x1 = x3;// f(x3)*f(x2)
					if (Math.abs(x2 - x3) < 0.00001) {
						m = (int) (x1 * 100);
						x1 = m / 100;
						System.out.println("方程的根是x=" + x1 + "，迭代次数=" + i);
						break;
					}

				}
			}
		}
	}

}