C程序 打印180度旋转的简单左半边金字塔

我们可以用for和while循环来打印简单的半左金字塔的180度旋转，如下所示。

输入:

rows = 5

输出:

    *
   **
  ***
 ****
*****

方法一

使用for循环

解释： 第一个for循环用来确定行的数量，第二个for循环用来确定列的数量。这里的值将根据第一个for循环而改变。如果j大于i，那么它将打印输出，否则打印空格。

// C program to print 180 degree rotation
// of a simple half left pyramid pattern
#include <stdio.h>
 
int main()
{
 
    int rows = 5;
 
    // first for loop is used to identify number of rows
    for (int i = rows; i > 0; i--) {
       
        // second for loop is used to identify number of
        // columns and here the values will be changed
        // according to the first for loop
        for (int j = 0; j <= rows; j++) {
           
            // if j is greater than i then it will print
            // the output otherwise print the space
            if (j >= i) {
                printf("*");
            }
            else {
                printf(" ");
            }
        }
        printf("\n");
    }
    return 0;
}

输出

     *
    **
   ***
  ****
 *****

方法二

使用while循环

解释： while 循环检查条件，直到条件为假。如果条件为真，则进入循环并执行这些语句。

// C program to print 180 degree rotation
// of a simple half left pyramid pattern
// using while loop
 
#include <stdio.h>
 
int main()
{
 
    int i = 0, j = 0, sp = 0;
    int rows = 5;
   
    // while loop check the condition until the given
    // condition is false if it is true then enteres in to
    // the loop
 
    while (i < rows) {
 
        // second while loop is used for printing spaces
        while (sp < (rows - i - 1)) {
            printf("  ");
            sp++;
        }
 
        // assigning sp value as 0 because we need to run sp
        // from starting
 
        sp = 0;
       
        // this loop will print the pattern
        while (j <= i) {
            printf("* ");
            j++;
        }
 
        j = 0;
        i++;
        printf("\n");
    }
    return 0;
}

输出

        * 
      * * 
    * * * 
  * * * * 
* * * * * 

时间的复杂性： 对于给定的输入n行，O(n 2)。

辅助空间： O(1)