Python & Java 數字金字塔解法

問題:讓電腦列印數字金字塔
            1
         1 2 1
      1 2 4 2 1
    1 2 4 8 4 2 1
 1 2 4 8 16 8 4 2 1 




 Java解法:
我以位移運算解決
1 2 4 8 16 8 4 2 1
問題 

int k = 5;

for (int i = 0; i < k; i++) {
 System.out.print((1 << i) + " ");
}

會打印 1 2 4 8 16

for (int j = 0; j < k+1; j++) {
 System.out.print(((1 << k) >> j) + " ");
}

再接著打印  8 4 2 1

int num = 5;
for (int k = 0; k < num; k++) {
 //printSpace(num-k);
 for (int i = 0; i < k; i++) {
  System.out.print((1 << i) + " ");
 }
 for (int j = 0; j < k+1; j++) {
  System.out.print(((1 << k) >> j) + " ");
 }
  System.out.print("\n");
 }
}

1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1

這樣子只有直角三角形，只要在打印每行前先打印二格空白"  "
就能呈現金字塔了
完整代碼如下

public class Pyramid {
 public static void main(String[] args) {
 int num = 5;
 for (int k = 0; k < num; k++) {
  printSpace(num-k);
  for (int i = 0; i < k; i++) {
   System.out.print((1 << i) + " ");
  }
  for (int j = 0; j < k+1; j++) {
   System.out.print(((1 << k) >> j) + " ");
  }
   System.out.print("\n");
  }
 }
 
 public static void printSpace(int num) {
  for (int s = 0; s < num; s++) {
   System.out.print("  ");
  }
 }
}

因為是先用python做測試，隨便附上python版

def pyramid(num):
    for i in range(num):
        print '\0\0'*(num-i),
        for j in range(i):
            print 1<> k,
        print