【学习ios之路:C语言】二维.三维数组.字符串数组的应用

一.二维数组

int a[3][4] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};//定义了一个3行4列的二维数组.

printf("%ld\n", sizeof(a));//48存储空间大小为48.

通过双下标访问二维数组中的元素,先访问行,再访问列.

注:二维数组定义时，可以不指定第⼀维的长度，但必须给定第二维。例如：int a[][3] = {1 , 2 , 3 , 4 , 5};

例子:

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int a[3][4] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
for (int i = 0; i < 3; i++) {
//外层循环控制行数;
for (int j = 0; j < 4; j++) {
//内层循环控制列数
printf("%4d ", a[i][j]); //输出二维数组的元素
}
printf("\n");
}

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2.定义一个二维数组,4行3列,元素取值范围为[20,40] 求所有元素的和.

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int n[4][3] = {0};
int  sum = 0;
for (int i = 0; i < 4; i++) {
for (int  j = 0; j < 3; j++) {
n[i][j] = arc4random() % ( 40 - 20 + 1) + 20;
sum  += n[i][j];
}
}
printf("%d\n", sum);

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3.输出每一行元素的和.

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int n[4][3] = {0};
int  sum = 0;
for (int i = 0; i < 4; i++) {
for (int  j = 0; j < 3; j++) {
n[i][j] = arc4random() % ( 40 - 20 + 1) + 20;
sum  += n[i][j];
printf("%d ", n[i][j]);
}
printf("sum = %d \n", sum);
sum = 0;
}

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4.输出每一列元素的和.

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int m[4][3] = {0};//定义二维数组
for (int i = 0 ; i < 4; i++) {
for (int j = 0; j < 3; j++) {
m[i][j] = arc4random() % (20 - 10 + 1 ) + 10; //随机值范围 10 - 20
printf("%d ", m[i][j]);
}
printf("\n");
}
//每一列的和
int sum = 0 ;
for (int i = 0; i < 3; i++) { //列
for (int j = 0; j < 4; j++) { //行
sum += m[j][i]; //m[j][i] //对应4行3列
}
printf("sum = %d ", sum);
sum  = 0;
}

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3、找出一个二维数组中的“鞍点”，即该位置上的元素在该行中最大，在该列中最小（也可能没鞍点），打印出有关信息。3行4列的二维数组,取值范围20 - 40

代码如下:

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//方法1:
//定义一个二维数组
int a[3][4] = {0};
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 4; j++) {
a[i][j] = arc4random() % (40 - 20 + 1) + 20;//随机取出几个范围在20-40的数.
printf("%d ", a[i][j]);
}
printf("\n");
}
//定义最大最小值
int max = 0, min = 1000;
int m[3] = {0};//定义用来存储最大值的一位数组.
int n[4] = {0};//定义用来存储最小值的一位数组.
//求最大值,并把最大值保存在一维数组m中.
for (int i = 0; i < 3; i++) {
for ( int j = 0; j < 4; j++) {
if (max < a[i][j]) {
max = a[i][j];
}
}
m[i] = max;
printf("max = %d ", max);
max = 0; //没遍历一次,max赋值为0
}
//求最小值,并把最小是保存在一维数组m中
for (int i = 0 ; i < 3; i++) {
for (int j = 0; j < 4; j++) {
if (min > a[i][j]) {
min = a[i][j];
}
}
n[i] = min;
printf("max = %d" , min);
min = 10000; //初始化
}
for (int i = 0 ; i < 3 ; i++) {
for (int j = 0; j < 4; j++) {
if (a[i][j] == m[i] && a[i][j] == n[j] ) {
printf("第%d行第%d列存在鞍点是:%d\n", i, j,a[i][j]);
}else {
printf("没有鞍点!");
 }
}

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方法2:

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int a[3][4] = {0};
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            a[i][j] = 1;//arc4random() % (40 - 20 + 1) + 20;
            printf("%d ", a[i][j]);
        }
        printf("\n");
    }
    //遍历数组中的每个元素,验证每个元素是否是鞍点.
    int count = 0; //用来记录鞍点的个数
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            //首先假定该数是该行中最大的
            BOOL isMax = YES;
            //验证该数是否是该行中最大的
            for (int column = 0; column < 4; column++) {
                if (a[i][column] > a[i][j]) {
                    isMax = NO; //如果不是最大的, 将isMax标识为NO.
                    break; //结束当前循环
                }
            }
            
            if (isMax == NO) {
                //如果为NO,说明不是该行中最大的,已经不满足鞍点的条件了.
                continue; //直接进入下一循环
            }
            //先假设是该列中最小的.
            BOOL isMin = YES;
            //验证该数是否是该列中最小的.
            for (int row = 0; row < 3; row++) {
                if (a[row][j] < a[i][j]) {
                    isMin = NO;
                    break;
                }
            }
            if (isMin == NO) {
                continue;
            }
            count++;
            printf("row:%d,column:%d = %d\n", i + 1, j + 1, a[i][j]);
        }
    }
    if (count == 0) {
        printf("没有鞍点");
    }

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