二维数组蛇形和Z形矩阵输出

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1、蛇形矩阵输出

思路：就顺时针的矩阵为例，由矩阵外至内圈的数值是逐渐加1的，所以主要是找出值循环增加的规律，跟设定的圈数round有关，而round的取值范围为0 <= round <= n / 2,首先定义圈数从0开始，→的下标x是恒定的，从0开始，但是下标y是从round至n - round的，接着向下↓，下标y保持n - round 不变，下标x从round + 1增加到n - round - 1,；再接着向左←，下标x保持n - round - 1不变，而下标y则由n - round - 1减到round，最后向上↑，保持下标y为round不变，再将下标x由n - round - 1减少到round + 1,一个圈结束后，round加1,继续循环，最后判断一下输入的矩阵函数是奇数还是偶数，如果是奇数就不能忘了矩阵最中间的数，最后输出的数。矩阵的各个数都已经设定好了，就以依次打印出矩阵。

#include <stdio.h>#include <stdlib.h>int main(){    int x;   /*二维数组的第一个下标*/    int y;   /*二维数组的第二个下标*/    int n;   /*创建的矩阵为n＊n*/    int count = 1;  /*蛇形矩阵由1开始计数*/    int round;     /*矩阵转的圈数*/        printf("Please input the size of the array you want to be(n * n):\n");    scanf("%d",&n);    int (*a)[n] = calloc(n * n,sizeof(int));    //如果n = 1,则直接输出    if(n == 1)    {        a[0][0] = count;    }    else    {        for(round = 0;round <= n / 2;round++)        {            x = round;            for(y = round;y < n - round;y++)            {                a[x][y] = count;                count++;            }                        y = n - round - 1;            for(x = round + 1;x < n - round - 1;x++)            {                a[x][y] = count;                count++;            }            x = n - round - 1;            for(y = n - round - 1;y >= round;y--)            {                a[x][y] = count;                count++;            }            y = round;            for(x = n - round - 2;x > round;x--)            {                a[x][y] = count;                count++;            }        }        if(n % 2 == 1)        {            a[n / 2][n / 2] = count;        }    }    for(x = 0;x< n;x++)    {        for(y = 0;y < n;y++)        {            printf("%3d",a[x][y]);        }        printf("\n");    }    printf("\n");        return 0;}

调试结果：

Please input the size of the array you want to be(n * n):5  1  2  3  4  5 16 17 18 19  6 15 24 27 20  7 14 23 22 21  8 13 12 11 10  9

逆时针打印：

#include <stdio.h>#include <stdlib.h>int main(){    int n;    int x;    int y;    int round;    int count = 1;    printf("Please input the size of array:\n");    scanf("%d",&n);    int (*a)[n] = calloc(n * n,sizeof(int));        if(n == 1)    {        a[0][0] = count;    }    else    {        for(round = 0;round < n / 2;round++)        {            y = round;            for(x = round;x < n - 1 - round;x++)            {                a[x][y] = count;                count++;            }            x = n - 1 - round;            for(y = round;y < n - 1 -round;y++)            {                a[x][y] = count;                count++;            }            y = n - 1 - round;            for(x = n - 1 - round;x > round;x--)            {                a[x][y] = count;                count++;            }            x = round;            for(y = n - 1 - round;y > round;y--)            {                a[x][y] = count;                count++;            }        }        if(n % 2 == 1)        {            a[n / 2][n / 2] = count;        }    }    for(x = 0;x < n;x++)    {        for(y = 0;y < n;y++)        {            printf("%3d",a[x][y]);        }        printf("\n");    }    printf("\n");        free(a);    return 0;}

调试结果：

Please input the size of array:5  1 16 15 14 13  2 17 24 23 12  3 18 25 22 11  4 19 20 21 10  5  6  7  8  9

2、Z型蛇形输出矩阵

思路：首先根据要输出的矩阵的大小决定一次增加的数组值的最大个数num即n和循环的次数round(2 * n - 1)，一次循环定义的二维数组值的下标x和y的和为round - 1，一直都是斜向上增加。

#include <stdio.h>#include <stdlib.h>int main(){     int x;    int y;    int n;    int num;    int round;    int count = 1;    printf("Please input the size of the array:\n");    scanf("%d",&n);    int (*a)[n] = calloc(n * n,sizeof(int));        if(1 == n)    {        a[0][0] = count;    }    else    {        for(round = 1;round <= n;round++)        {            for(y = 0;y < round;y++)            {                a[round - 1 - y][y] = count;                count++;            }        }        for(;round <= 2 * n - 1;round++)        {            for(y = round - n;y < n;y++)            {                a[round - 1 - y][y] = count;                count++;            }        }    }    for(x = 0;x < n;x++)    {        for(y = 0;y < n;y++)        {            printf("%3d",a[x][y]);        }        printf("\n");    }    printf("\n");    free(a);    return 0;}

调试结果：

Please input the size of the array:7  1  3  6 10 15 21 28  2  5  9 14 20 27 34  4  8 13 19 26 33 39  7 12 18 25 32 38 43 11 17 24 31 37 42 46 16 23 30 36 41 45 48 22 29 35 40 44 47 49

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