Java解决 过河问题

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Test3 {
static int sum = 0; // 存放总共花费的时间
static List record = new ArrayList(); // 存放每趟用去的时间
// 思路：从大数开始运送，小数仅充当“邮递员”。
static void go(int[] a) {
int n = a.length;
int sum1, sum2;
int i;
for (i = n - 1; i > 2; i = i - 2) {

sum1 = a[1] + (a[0]) + a[i] + a[1];
sum2 = a[i] + a[0] + a[i - 1] + a[0];
if (sum1 < sum2) {
sum = sum + sum1;

record.add(a[1]);
record.add(a[0]);
record.add(a[i]);
record.add(a[1]);
System.out.println("(" + a[0] + "," + a[1] + ")" + "去，" +
a[0]
+ "回； " + "(" + a[i - 1] + "," + a[i] + ")去," + a[1]
+ "回；");

} else {
sum = sum + sum2;

record.add(a[i]);
record.add(a[0]);
record.add(a[i - 1]);
record.add(a[0]);
System.out.println("(" + a[0] + "," + a[i] + ")去," +
a[0]
+ "回； " + "(" + a[0] + "," + a[i - 1] + ")去," + a[0]
+ "回；");
}
}

if (i == 2) {
sum = sum + a[0] + a[1] + a[2]; // 只剩三个数时，时间为三者之和

record.add(a[0]);
record.add(a[1]);
record.add(a[2]);
System.out.println("(" + a[0] + "," + a[2] + ")去," + a[0] +
"回； "
+ "(" + a[0] + "," + a[1] + ")去。" + "完成！");
} else if (i == 1) {
sum = sum + a[1]; // 只剩两个数时，时间取大

record.add(a[1]);
System.out.println("(" + a[0] + "," + a[1] + ")去。" +
"完成！");
}

}

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();

int i;

int[] a = new int
; // a[] 已排序

for (i = 0; i < n; i++) {
a[i] = sc.nextInt();
}

go(a);

for (i = 0; i < record.size() - 1; i++) {
System.out.print(record.get(i) + "+");
}
System.out.print(record.get(i));

System.out.println("=" + sum);

}

}