24点算法的Java和C语言源码.docVIP

24点算法的Java版和C语言版源码 Java版共2个类：Application.java 和 Node.java == Application.java== package num24; public class Application { double num[] = new double[4]; boolean mark[] = new boolean[4]; /** * 获得k个数运算，结果为result的第一个Node，比如get(24,4) * @param result * @param n * @return */ public Node get(double result, int n) { Node left = null, right = null; if (n = 0) { return null; } //如果个数小于0，返回空 for (int i = 0; i 4; i++) { if (mark[i]) { continue; //如果已经打上标志，继续找下一个 } mark[i] = true; //如果没有打标志，就打上标志，并且对其进行处理 //如果只有一个数并且这个数就是给出的结果，就返回一个只有一个数的节点 if (result == num[i] n == 1) { return new Node(num[i]); } left = get(result - num[i], n - 1); if (left != null) { return new Node(0.0, +, left, new Node(num[i])); } left = get(result + num[i], n - 1); if (left != null) { return new Node(0.0, -, left, new Node(num[i])); } left = get(result * num[i], n - 1); if (left != null) { return new Node(0.0, /, left, new Node(num[i])); } if (Math.abs(num[i]) 1e-10) { left = get(result / num[i], n - 1); if (left != null) { return new Node(0.0, *, left, new Node(num[i])); } } for (int j = 0; j 4; j++) { if (mark[j]) { continue; } mark[j] = true; if (result == num[i] * num[j] n == 2) { //a*b return new Node(0.0, *, new Node(num[i]), new Node(num[j])); } left = get(result + num[i] * num[j], n - 2); //a*b-(c?d)=result if (left != null) { right = new Node(0.0, *, new Node(num[i]), new Node(num[j])); return new Node(0.0, -, left, right); } left = get(result - num[i] * num[j], n - 2); //a*b+(c?d)=result if (left != null) { right = new Node(0.0, *, new Node(num[i]), new Node(num[j])); return new Node(0.0, +, left, right); } if (Math.