数据结构
网络
关系型数据库管理系统
操作系统
Java
MS Excel
iOS
HTML
CSS
Android
Python
C 编程
C++
C#
MongoDB
MySQL
Javascript
PHP
物理学
化学
生物学
数学
英语
经济学
心理学
社会学
时尚学
法律研究以 Z 字形方式打印 C 编程中的矩阵。
给定一个矩阵 mat[row][col]，我们必须按照如下给出的图像以 Z 字形方式打印给定的矩阵-
所以输出应该是这样的 -
Output: 10 20 40 70 50 30 60 80 90
对于以上问题，我们遵循一种简单的方法，其中我们必须对角迭代矩阵，并在每次以前匹配项后更改迭代值以改变方向。
算法
START STEP 1-> DECALRE AND SET k = 3, l = 3 STEP 2-> DECLARE A MATRIX mat[][3] STEP 3-> DECLARE AND SET row = 0, col = 0, flag = false; STEP 4-> SET mn = MINIMUM(k, l) STEP 5-> LOOP FOR len = 1 AND len <= mn AND ++len LOOP FOR i = 0 AND i < len AND ++i PRINT mat[row][col] IF i + 1 == len THEN, BREAK END IF IF flag THEN, INCREMENT row BY 1 AND DECREMENT col BY 1 ELSE DECREMENT row BY 1 AND INCREMENT col BY 1 END IF END FOR IF len == mn THEN, BREAK END IF IF flag THEN INCREMENT row BY 1 AND SET flag = FALSE ELSE INCREMENT col BY 1 AND SET flag = TRUE END IF END FOR STEP 6-> IF row == 0 THEN, IF col == k – 1 THEN, INCREMENT row BY 1 ELSE INCREMENT col BY 1 END IF SET flag = 1 ELSE IF row == l – 1 THEN, INCREMENT col BY 1 ELSE INCREMENT row BY 1 SET flag = 0 END IF STEP 7-> SET MAX = MAXIMUM(k, l) – 1 STEP 8-> LOOP FOR len, diag = MAX AND diag > 0 AND --diag IF diag > mn THEN, SET len = mn ELSE SET len = diag END IF FOR i = 0 AND i < len AND ++i PRINT mat[row][col] IF i + 1 == len THEN, BREAK END IF IF flag THEN, INCREMENT row BY 1 AND DECREMENT col BY 1 ELSE INCREMENT col BY 1 AND DECREMENT row BY 1 END IF IF row == 0 || col == k – 1 THEN, IF col == k - 1 INCREMENT row BY 1 ELSE INCREMENT col BY 1 END IF SET flag = true ELSE IF col == 0 || row == l – 1 THEN, IF row == l – 1 THEN, INCREMENT col BY 1 ELSE INCREMENT row BY 1 END IF SET flag = false ENF IF END FOR STOP
示例
#include <stdio.h>
#include <stdbool.h>
#define C 3
#define min(a, b) a>b?b:a
#define max(a, b) a>b?a:b
int main(){
int k = 3, l = 3;
int mat[][3] = {
{ 10, 20, 30 },
{ 40, 50, 60 },
{ 70, 80, 90 }
};
int row = 0, col = 0;
bool flag = false;
int i, j, len, diag;
int MAX;
int mn = min(k, l); //to check the minimum number and return that minimum number
for ( len = 1; len <= mn; ++len) {
for ( i = 0; i < len; ++i) {
printf("%d ", mat[row][col]); //Printing the matrix in zigzag format
if (i + 1 == len)
break;
if (flag)
++row, --col;
else
--row, ++col;
}
if (len == mn)
break;
if (flag)
++row, flag = false;
else
++col, flag = true;
}
if (row == 0) {
if (col == k - 1)
++row;
else
++col;
flag = 1;
} else {
if (row == l - 1)
++col;
else
++row;
flag = 0;
}
MAX = max(k, l) - 1; //To check the maximum element
for ( len, diag = MAX; diag > 0; --diag) { //Loop to go diagonally.
if (diag > mn)
len = mn;
else
len = diag;
for ( i = 0; i < len; ++i) {
printf("%d ", mat[row][col]);
if (i + 1 == len)
break;
if (flag)
++row, --col;
else
++col, --row;
}
if (row == 0 || col == k - 1) {
if (col == k - 1)
++row;
else
++col;
flag = true;
}
else if (col == 0 || row == l - 1) {
if (row == l - 1)
++col;
else
++row;
flag = false;
}
}
return 0;
}输出
此程序将打印输出 -
10 20 40 70 50 30 60 80 90
更新于： 04-Sep-2019
1000+ 次浏览
广告