题目:将一个7*7(只能要求必须行和列相等的矩阵)的二维数组下三角部分以螺旋、依次+1的方式填满并输出(如下图所示)
方法:依旧设置四个变量Left、Right、Up、Down控制循环,不过只有三种模式,故只需要三个for循环(红色线条、蓝色线条、绿色线条)即可解决。
代码如下:(需要注意的已在代码中用注释说明)
#include<stdio.h>
void Print_Ar(int br[7][7],int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%3d", br[i][j]);
}
printf("\n");
}
printf("\n");
}
int main() {
const int n = 7;
int ar[n][n] = { 0 };//该代码只适用于n*n的二维数组下三角填充!!
int Left = 0, Right = n - 1, Up = 0, Down = n - 1;
int num = 1;
while (Left <= Right && Up <= Down) {
//需要填充三个不同部分,故需使用三个循环解决
for (int i = Up; i <= Down; i++) {
ar[i][Left] = num++;
}
for (int i = Left + 1; i <= Right; i++) {
ar[Down][i] = num++;
}
//切勿写成下列循环!!只有第一次填充是对角线ar[i][i]!!
/*for (int i = Down - 1; i > Up; i--) {
ar[i][i] = num++;
}*/
for (int i = Down - 1, j = Right - 1; i > Up&& j > Left; i--, j--) {
ar[i][j] = num++;
}
//不要忘记控制循环变量发生改变
Left++, Right -= 2, Up += 2, Down--;//通过观察后续循环发现的规律
}
Print_Ar(ar, n);
return 0;
}
运行结果如下:
同理,上三角填充也很容易。如下图所示:
代码如下:
#include<stdio.h>
void Print_Ar(int br[7][7], int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%3d", br[i][j]);
}
printf("\n");
}
printf("\n");
}
int main() {
const int n = 7;
int ar[n][n] = { 0 };
int left = 0, right = n - 1, up = 0, down = n - 1;
int num = 1;
while (left <= right && up <= down) {
for (int i = left; i <= right; i++) {
ar[up][i] = num++;
}
for (int i = up + 1; i <= down; i++) {
ar[i][right] = num++;
}
for (int i = down - 1, j = right - 1; i > up && j > left; i--, j--) {
ar[i][j] = num++;
}
up++, down -= 2, left += 2, right--;
}
Print_Ar(ar, n);
return 0;
}
运行结果如下: