一道循环题

Question

 打印一个 N*N 的方阵，Ｎ为每边                  Ｎ＝１５  打印出下面图形
 字符的个数（３＜Ｎ＜２０）, 要求最              TTTTTTTTTTTTTTT
 外一层为"T", 第二层为"J", 从第三层             TJJJJJJJJJJJJJT
 起每层依次打印数字 1,2,3,...                 TJ11111111111JT
 （右图以Ｎ为１５为例）                           TJ12222222221JT
                                            TJ12333333321JT
                                            TJ12344444321JT
                                            TJ12345554321JT
                                            TJ12345654321JT
                                            TJ12345554321JT
                                            TJ12344444321JT
                                            TJ12333333321JT
                                            TJ12222222221JT
                                            TJ11111111111JT
                                            TJJJJJJJJJJJJJT
                                            TTTTTTTTTTTTTTT

Answer 1

Ｎ＝１５  打印出下面图形 TTTTTTTTTTTTTTT 
TJJJJJJJJJJJJJT 
TJ11111111111JT 
TJ12222222221JT 
TJ12333333321JT 
TJ12344444321JT 
TJ12345554321JT 
TJ12345654321JT 
TJ12345554321JT 
TJ12344444321JT 
TJ12333333321JT 
TJ12222222221JT 
TJ11111111111JT 
TJJJJJJJJJJJJJT 
TTTTTTTTTTTTTTT
帮排版

Answer 2

/**
 * 算法说明：
 * 分析该矩阵需求，有一个特点，除去最外层两圈，
 * 中间所打印的数字的规律，可以认为是该数字所在位置距离四个边的距离中最短的，
 * 再仔细分析，如果将J和T也算作数字，也能符合这个规律（假定J是0，T是-1)。
 * 
 * 所以，问题简化为，算出某一个位置据四边的最短距离，然后取“TJ123456789”这个字符串中响应位置的字符打印即可。
 * 
 * 
 * @author jinxfei
 *
 */
public class Test {
private static String allChars="TJ123456789";
public static void main(String[] args){

int order=20;//矩阵的大小

for(int i=0; i<=order;i++){
for (int j=0; j<=order;j ++){
System.out.print(getChar(j,i,order));
}
System.out.println();
}

}

//计算当前坐标应该打印的字符
private static char getChar(int y,int x, int order){
//首先算出当前坐标据四个边的位置
int toLeft=y;
int toTop=x;
int toRight=order-y;
int toBottom=order-x;
//取其中的最小值
return allChars.charAt(getMin(toLeft,toTop,toRight,toBottom));
}

//计算取四个数中最小的值
private static int getMin(int a1, int a2, int a3, int a4){
int min=getMin(a1,a2);
min=getMin(min,a3);
min=getMin(min,a4);
return min;
}
//取两个数字中的最小值。
private static int getMin(int a1, int a2){
int min=a1;
if (min>a2){
min=a2;
}
return min;
}
}