java判断4个10以内的数字，通过四则运算得到24
时间:2011-02-05 赵学庆
请你用java写一个方法，判断给定的4个数字，每个数值在1~10之间，通过+ - * / 运算，结果为24，每个数字必须使用且只能使用一次。
网友提供的穷举法
public class T {
int[] can = new int[4];
public static void main(String[] args) {
T tf = new T(3, 4, 5, 7);
System.out.println(tf.getResult());
}
public T(int a, int b, int c, int d) {
can[0] = a;
can[1] = b;
can[2] = c;
can[3] = d;
}
public String getResult() {
double result1 = 0;
double result2 = 0;
double result3 = 0;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
for (int k = 0; k < 4; k++) {
if (i != j) {
result1 = cal(can[i], can[j], k);
for (int l = 0; l < 4; l++) {
for (int m = 0; m < 4; m++) {
if (i != l && j != l) {
result2 = cal(result1, can[l], m);
for (int n = 0; n < 4; n++) {
for (int o = 0; o < 4; o++) {
if (i != n && j != n && l != n) {
result3 = cal(result2, can[n], o);
if (result3 == 24) {
return result(can[i], k, can[j], m, can[l], o, can[n]);
}
}
}
}
}
}
}
}
}
}
}
return null;
}
private String result(int can1, int op1, int can2, int op2, int can3, int op3, int can4) {
return "" + can1 + convert(op1) + can2 + convert(op2) + can3 + convert(op3) + can4;
}
private String convert(int opr) {
if (opr == 0) {
return "+";
} else if (opr == 1) {
return "-";
} else if (opr == 2) {
return "*";
} else if (opr == 3) {
return "/";
}
return "";
}
private double cal(double a, double b, int opr) {
if (opr == 0) {
return a + b;
} else if (opr == 1) {
return a - b;
} else if (opr == 2) {
return a * b;
} else if (opr == 3) {
return a / b;
}
return 0;
}
}
运算结果
3*4+5+7 |
