二维数组类编程题

最大值

  1. 定义一个5*5的二维数组并初始化，找出数组中每行的最大值，放入一个新的数组。
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int main()
{
    int arr1[5][5] = {
        {1, 6, 3, 7, 5},
        {34, 4235, 5, 32, 4},
        {354, 435, 34, 23, 54},
        {325, 56, 98, 454, 3},
        {3, 5, 7, 6, 4},
    };
    int arr2[5] = {0};
    int i, j;
    for (i = 0; i < 5; i++)
    {
        for (j = 0; j < 5; j++)
        {
            if (arr1[i][j] > arr2[i])
            {
                arr2[i] = arr1[i][j];
            }
        }
    }

    for (i = 0; i < 5; i++)
    {
        printf("%d ", arr2[i]);
    }

    return 0;
}
  1. 定义一个5*5的二维数组并初始化，找出数组中的最大值的坐标。
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int main()
{
    int arr[5][5] = {
        {1, 6, 3, 7, 5},
        {34, 4235, 5, 32, 4},
        {354, 435, 34, 23, 54},
        {325, 56, 98, 454, 3},
        {3, 5, 7, 6, 4},
    };
    int i, j;
    int max_i = 0, max_j = 0;
    for (i = 0; i < 5; i++)
    {
        for (j = 0; j < 5; j++)
        {
            if (arr[i][j] > arr[max_i][max_j])
            {
                max_i = i;
                max_j = j;
            }
        }
    }
    printf("%d", arr[max_i][max_j]);

    return 0;
}

对角线

  1. 判断 a[N][N] 是否关于主对角线对称(左斜),若对称输出 YES,否则输出 NO。

主对角线：从左上角到右下角

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#include <stdio.h>
#define N 5

int main()
{
    int arr[N][N] = {
        {1, 2, 3, 4, 5},
        {2, 1, 3, 2, 4},
        {3, 3, 1, 0, 3},
        {4, 2, 0, 1, 2},
        {5, 4, 3, 2, 1},
    };
    int i, j;
    int flag = 1;
    for (i = 0; i < N; i++) // i=0,1,2,3,4
    {
        for (j = i + 1; j < N; j++) // j=1,2,3,4
        {
            if (arr[i][j] != arr[j][i])
            {
                flag = 0;
                break;
            }
        }
        if(flag == 0)
        {
            break;
		}
    }
    if (flag == 1)
    {
        printf("YES");
    }
    else
    {
        printf("NO");
    }

    return 0;
}
  1. 判断 a[N][N]是否关于次对角线对称(右斜),若对称输出 YES,否则输出 NO。
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int main()
{
    int arr[N][N] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20},
        {21, 22, 23, 24, 25},
    };
    int i, j;
    int flag = 1;
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < N - i - 1; j++)
        {
            if (arr[i][j] != arr[N - j - 1][N - i - 1])
            {
                flag = 0;
                break;
            }
        }
        if (flag == 0)
        {
            break;
        }
    }
    if (flag == 1)
    {
        printf("YES");
    }
    else
    {
        printf("NO");
    }

    return 0;
}
  1. 分别求出 N 阶方阵 a 中两个对角线上元素之和。
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int main()
{
    int arr[N][N] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20},
        {21, 22, 23, 24, 25},
    };
    int i, j;
    int sum1 = 0, sum2 = 0;

    for (i = 0; i < N; i++)
    {
        sum1 = sum1 + arr[i][i];
        sum2 = sum2 + arr[i][N - i - 1];
    }
    printf("%d\n", sum1);
    printf("%d\n", sum2);

    return 0;
}
  1. 定义一个5*5的二维数组并初始化，分别计算出两条对角线的和各为多少。

主对角线：arr[0][0]、arr[1][1]、arr[2][2]、arr[3][3]

辅对角线：arr[0][3]、arr[1][2]、arr[2][1]、arr[3][0]

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方法1：
int main()
{
    int arr[N][N] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20},
        {21, 22, 23, 24, 25},
    };
    int i, j;
    int sum1 = 0, sum2 = 0;
    for (i = 0; i < N; i++)
    {
        sum1 += arr[i][i];
        sum2 += arr[i][N - i - 1];
    }

    printf("sum1 = %d\nsum2 = %d", sum1, sum2);

    return 0;
}

方法2：
int main()
{
    int arr[N][N] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20},
        {21, 22, 23, 24, 25},
    };
    int i, j;
    int sum1 = 0, sum2 = 0;
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < N; j++)
        {
            if (i == j)
            {
                sum1 += arr[i][j];
            }
            if (i + j == N)
            {
                sum2 += arr[i][j];
            }
        }
    }
    printf("sum1 = %d\nsum2 = %d", sum1, sum2);

    return 0;
}

反转

  1. 上下对称反转
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#define ROW 4
#define COL 5

int main()
{
    int arr[ROW][COL] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20},
    };
    int tmp;
    int i, j;
    for (i = 0; i < ROW / 2; i++)
    {
        for (j = 0; j < COL; j++)
        {
            tmp = arr[i][j];
            arr[i][j] = arr[ROW - 1 - i][j];
            arr[ROW - 1 - i][j] = tmp;
        }
    }
    for (i = 0; i < ROW; i++)
    {
        for (j = 0; j < COL; j++)
        {
            printf("%d\t", arr[i][j]);
        }
        printf("\n");
    }
    
    return 0;
}
输出：
16      17      18      19      20
11      12      13      14      15
6       7       8       9       10
1       2       3       4       5
  1. 左右对称反转
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#define ROW 4
#define COL 5

int main()
{
    int arr[ROW][COL] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20},
    };
    int tmp;
    int i, j;

    for (i = 0; i < ROW; i++) // i=0,1,2,3
    {
        for (j = 0; j < COL / 2; j++) // j=0,1
        {
            tmp = arr[i][j];
            arr[i][j] = arr[i][COL - j - 1];
            arr[i][COL - j - 1] = tmp;
        }
    }

    for (i = 0; i < ROW; i++)
    {
        for (j = 0; j < COL; j++)
        {
            printf("%d\t", arr[i][j]);
        }
        printf("\n");
    }
}
输出：
5       4       3       2       1
10      9       8       7       6
15      14      13      12      11
20      19      18      17      16
  1. 上下左右反转
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#define ROW 4
#define COL 5

int main()
{
    int arr[ROW][COL] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15},
        {16, 17, 18, 19, 20},
    };
    int tmp;
    int i, j;
    // 上下对称反转
    for (i = 0; i < ROW / 2; i++)
    {
        for (j = 0; j < COL; j++)
        {
            tmp = arr[i][j];
            arr[i][j] = arr[ROW - i - 1][j];
            arr[ROW - i - 1][j] = tmp;
        }
    }
    // 左右对称反转
    for (i = 0; i < ROW; i++) // i=0,1,2,3
    {
        for (j = 0; j < COL / 2; j++) // j=0,1
        {
            tmp = arr[i][j];
            arr[i][j] = arr[i][COL - j - 1];
            arr[i][COL - j - 1] = tmp;
        }
    }
    // 打印
    for (i = 0; i < ROW; i++)
    {
        for (j = 0; j < COL; j++)
        {
            printf("%d\t", arr[i][j]);
        }
        printf("\n");
    }

    return 0;
}
输出：
20      19      18      17      16
15      14      13      12      11
10      9       8       7       6
5       4       3       2       1

杨辉三角

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#define N 10
int main()
{
    int arr[N][N] = {0};
    int i, j;
    for (i = 0; i < N; i++)
    {
        for (j = 0; j <= i; j++)
        {
            if (i == j || j == 0)
            {
                arr[i][j] = 1;
            }
            else
            {
                arr[i][j] = arr[i - 1][j] + arr[i - 1][j - 1];
            }
        }
    }

    for (i = 0; i < N; i++)
    {
        for (j = 0; j <= i; j++)
        {
            printf("%d ", arr[i][j]);
        }
        printf("\n");
    }

    return 0;
}

斐波那契

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#define N 20
int main()
{
    int arr[N] = {1, 1};
    int i = 0;
    for (i = 2; i < N; i++)
    {
        arr[i] = arr[i - 1] + arr[i - 2];
    }
    for (i = 0; i < N; i++)
    {
        printf("%d ", arr[i]);
    }

    return 0;
}

元素互换

将一个二维数组的行和列的元素互换，并输出。

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int main()
{
    int arr1[2][3] = {1, 2, 3, 4, 5, 6};
    int arr2[3][2] = {0};
    int i, j;
    for (i = 0; i < 2; i++)
    {
        for (j = 0; j < 3; j++)
        {
            printf("%d ", arr1[i][j]);
            arr2[j][i] = arr1[i][j];
        }
        printf("\n");
    }
    printf("\n");

    for (i = 0; i < 3; i++)
    {
        for (j = 0; j < 2; j++)
        {
            printf("%d ", arr2[i][j]);
        }
        printf("\n");
    }

    return 0;
}

输出：
1 2 3 
4 5 6

1 4
2 5
3 6