2个算法 求解！

Question

第一个：
设计一个三角螺旋输出如下：
12 93 10 84 5  6  7 第二个：
当N=4时能输出以下数字， 其中N为通用的，当N=任何一个数时都能输出以下规律的图形。
1     2     3     4
11    9     10    5
15    12    8     6
16    14    13    7第二个 是蛇形三角+螺旋三角拼接成的矩形，11-12-13 这条斜线是蛇形  1-9-8-7是螺旋。不知道如何写啊 各位 给出详细代码和注释啊。

Answer 1

输出下三角矩阵螺旋数，比如，输入为4，则输出：1 
2 9 
3 10 8 
4 5 6 7 
输入为5，则输出：1 
2 12 
3 13 11 
4 14 15 10 
5 6 7 8 9 
代码如下，大致思想就是设一个方向变量，-1为向下，0为向右，1为向左上，然后控制二维数组下标为数组赋值，最后打印。修改print()方法的传入参数即可。
view plaincopy to clipboardprint?public class Sort {  
    public static void main(String[] args){  
        print(5);  
    }  
    static void print(int n){  
        int l = n;  
        int sum = 0;  
        for(int i=1; i<=n; i++){  
            sum += i;  
        }  
        int a[][] = new int[n][n];  
        int s = 1;  
        int i = 0, j = 0;  
        int d = -1;  
        for(int k=1; k<=sum; k++){  
            switch (d) {  
            case -1:  
                a[i][j] = s++;  
                if(i<n-1){  
                    i++;  
                }  
                else{  
                    n--;  
                    j++;  
                    d = 0;  
                }  
                break;  
            case 0:  
                a[i][j] = s++;  
                if(i == (l-1)){  
                    if(j<i){  
                        j++;  
                    }  
                    else{  
                        i--;  
                        j--;  
                        d = 1;  
                    }  
                }  
                else if(a[i][j+1] == 0){  
                    j++;  
                }  
                else{  
                    i--;  
                    j--;  
                    d = 1;  
                }  
                break;  
            case 1:  
                a[i][j] = s++;  
                if(a[i-1][j-1] == 0){  
                    i--;  
                    j--;  
                }  
                else{  
                    i++;  
                    d = -1;  
                }  
                break;  
            default:  
                break;  
            }  
        }  
        for(int p=0; p<l; p++){  
            for(int q=0; q<=p; q++){  
                System.out.print(a[p][q] + " ");  
            }  
            System.out.println();  
        }  
    }  
}  

Answer 2

看看可以不            int n = 6;
            int[,] arr = new int[n, n];
            arr[0, 0] = 1;
            int X=0,Y=0;
            for (int i = 0; i < n; i++)
            {
                for (int k = 0; k < n - i; k++)
                {
                    if (i % 3 == 1)
                    {
                        arr[X, Y] = arr[X, Y - 1] + 1;
                        if (k == n - i - 1)
                        {
                            X--;
                            Y--;
                            break;
                        }
                        Y++;
                    }
                    if (i % 3 == 0)
                    {
                        if (X - 1 < 0)
                        {
                            X++;
                            continue;
                        }
                        arr[X, Y] = arr[X - 1, Y] + 1;
                        if (k == n - i - 1)
                        {
                            Y++;
                            break;
                        }
                        X++;
                    }
                    if (i % 3 == 2)
                    {
                        if (X < 0 || Y < 0)
                            continue;
                        arr[X, Y] = arr[X + 1, Y + 1] + 1;
                        if (k == n - i - 1)
                        {
                            X++;
                            break;
                        }
                        X--;
                        Y--;
                    }
                }
            }

Answer 3

先说第一个，第二个下周上班了再弄。            int size = 10;            int i=1;
            int x = 0;
            int y = 0;
            int[,] A = new int[size, size];            for (int j = 0; j < size; j++)
            {
                for (int k = 0; k < size; k++)
                {
                    A[j, k] = 0;
                }
            }            int flag = size * (size + 1) / 2;            while (i <= flag)
            {
                if (x + 1 < size)
                {
                    if (A[x + 1, y] == 0)
                    {
                        A[x, y] = i;
                        i++;
                        x++;
                        continue;
                    }
                }                if (y + 1 < size)
                {
                    if (y <= x)
                    {
                        if (A[x, y + 1] == 0)
                        {
                            A[x, y] = i;
                            i++;
                            y++;
                            continue;
                        }
                    }
                }                while (A[x, y] == 0)
                {
                    A[x, y] = i;
                    i++;
                    if (A[x - 1, y - 1] == 0)
                    {
                        x--;
                        y--;
                    }
                    else
                    {
                        x++;
                        break;
                    }
                }
            }            for (int j = 0; j < size; j++)
            {
                for (int k = 0; k <= j; k++)
                {
                    Console.Write("{0}\t", A[j,k]);
                    if (j == k)
                    {
                        Console.WriteLine();
                    }
                }
            }            Console.ReadLine();