我来贴一个，好像比1楼的短一点，不过自己依然不是很满意 int width = 6; int bits = 2; //每个数的宽度，为了对齐美观 String[] result = new String[width]; for (int i=0; i<width; i++) result[i] = ""; int current = 0; int direction = 1; int align ; //起始行 for (int i=1; i<2*width; i++){ int deep = width - Math.abs(i-width); if (i<=width) if (deep%2==1) align = 0; else align = deep - 1 ; else if (deep%2==0) align = width - 1; else align = width - deep; for (int j=1; j<=deep; j++){ current++; result[align] += String.format("%"+bits+"d", current) + " "; align += direction; } direction *= -1; } for (String s : result) System.out.println(s);
public static void main(String[] args) {
int n = 5; //邊長
int dn = n - 1;
int[][] val = new int[n][n];
for (int i = 0; i < n; i++) {
val[i] = new int[n];
}
int ct = 1;
boolean xTOy = true;
int x = 0;
int y = 0;
for (int i = -dn; i <= dn; i++) {
for (int j = dn - Math.abs(i); j >= 0; j--) {
val[y][x] = ct;
if (j > 0) {
if (xTOy) {
x--;
y++;
} else {
y--;
x++;
}
}
ct++;
}
if (x == 0) {
if (y < dn) {
y++;
} else {
x++;
}
} else if (y == 0) {
if (x < dn) {
x++;
} else {
y++;
}
} else if (y == dn) {
x++;
} else if (x == dn) {
y++;
}
if (xTOy) {
xTOy = false;
} else {
xTOy = true;
}
}
for (int i = 0; i < val.length; i++) {
for (int j = 0; j < val[i].length; j++) {
System.out.print((val[i][j] > 9 ? " " : " ") + val[i][j]);
}
System.out.println();
}
}
因為是斜的增減，所以以斜邊為軸，但這樣沒辦法直接列印，所以先寫到陣列裡。
以座標觀念來看
從0,0 開始 碰到邊界 不是Y+1就是X+1
然後每個下一點不是X-- Y++ 就是X++ Y--
以這樣的觀念去修修改改就可以了。
int bits = 2; //每个数的宽度，为了对齐美观
String[] result = new String[width];
for (int i=0; i<width; i++) result[i] = "";
int current = 0;
int direction = 1;
int align ; //起始行
for (int i=1; i<2*width; i++){
int deep = width - Math.abs(i-width);
if (i<=width)
if (deep%2==1) align = 0;
else align = deep - 1 ;
else
if (deep%2==0) align = width - 1;
else align = width - deep;
for (int j=1; j<=deep; j++){
current++;
result[align] += String.format("%"+bits+"d", current) + " ";
align += direction;
}
direction *= -1;
}
for (String s : result) System.out.println(s);