二分查找算法——Java版

binyan17

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;}}}}}}