转圈打印二维数组

题目要求：
思路：
我的最初思路：
将矩阵的四个边作为边界，一个循环执行四次九十度转弯，用swich、case实现，当四个边界指向一个位置结束循环，但是在拐点的处理上比较麻烦。
资料参考的思路：
写一个函数，输出二维数组一圈的元素。
循环执行上个函数，也就是输出每个圈的元素。
这个思路更好理解一些。
代码：

#include<iostream>
using namespace std;

void printEdge(int * m, int tR, int tC, int dR, int dC,int row,int col) {
    if (tR == dR) {

        // 子矩阵只有一行时
        for (int i = tC; i <= dC; i++) {
            cout<<*(m+i)<<endl;
        }


    } 

    else if (tC == dC) { 

        // 子矩阵只有一列时
        for (int i = tR; i <= dR; i++) {
            cout<<*(m+col*i+tC)<<endl;
        }


    } 

    else { // 一般情况
        int curC = tC;
        int curR = tR;
        while (curC != dC) {
            cout<<*(m+tR*col+curC)<<endl;
            curC++;
        }
        while (curR != dR) {
            cout<<*(m+curR*col+dC)<<endl;
            curR++;
        }
        while (curC != tC) {
            cout<<*(m+col*dR+curC)<<endl;
            curC--;
        }
        while (curR != tR) {
            cout<<*(m+col*curR+tC)<<endl;
            curR--;
        }
    }
}



void spiralOrder(int* matrix,int row,int col)
{
    int tR = 0;
    int tC = 0;
    int dR = row -1;
    int dC = col -1;
    while(tR<=dR&&tC<=dC)
    {
        printEdge(matrix, tR++, tC++, dR--, dC--,col,row);
    }
}




int main()
{

    int a[][4] = {1,2,3,6};
    spiralOrder((int*) a,1,4);

    int b[][4] = {
        {1},
        {2},
        {3},
        {16}
    };
    spiralOrder((int*) b,4,1);
    cout<<"/////////////////////////"<<endl;
    int c[][4] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
    spiralOrder((int*) c,4,4);



/* int N = 4; int sum = 0; int flag = 0; int right = N-1; int left = 0; int up = 1; int down = N-1; int x = 0; int y = -1; while(sum<N*N) { switch(flag) { case 0: y = y+1; while(y<=right) { cout<<a[x][y]<<endl; sum = sum +1; y++; } y = right; right = right-1; flag = 1; break; case 1: x = x+1; while(x<=down) { cout<<a[x][y]<<endl; sum = sum +1; x++; } x = down; down = down-1; flag = 2; break; case 2: y = y-1; while(y>=left) { cout<<a[x][y]<<endl; y = y-1; sum++; } y = left; left = left +1; flag = 3; break; case 3: x = x-1; while(x>=up) { cout<<a[x][y]<<endl; x = x-1; sum++; } x = up; up = up+1; flag = 0; break; default: break; } }*/

    return 0;
}

java：

package chapter_8_arrayandmatrix;

public class Problem_01_PrintMatrixSpiralOrder {

    public static void spiralOrderPrint(int[][] matrix) {
        int tR = 0;
        int tC = 0;
        int dR = matrix.length - 1;
        int dC = matrix[0].length - 1;
        while (tR <= dR && tC <= dC) {
            printEdge(matrix, tR++, tC++, dR--, dC--);
        }
    }

    public static void printEdge(int[][] m, int tR, int tC, int dR, int dC) {
        if (tR == dR) { // 子矩阵只有一行时
            for (int i = tC; i <= dC; i++) {
                System.out.print(m[tR][i] + " ");
            }
        } else if (tC == dC) { // 子矩阵只有一列时
            for (int i = tR; i <= dR; i++) {
                System.out.print(m[i][tC] + " ");
            }
        } else { // 一般情况
            int curC = tC;
            int curR = tR;
            while (curC != dC) {
                System.out.print(m[tR][curC] + " ");
                curC++;
            }
            while (curR != dR) {
                System.out.print(m[curR][dC] + " ");
                curR++;
            }
            while (curC != tC) {
                System.out.print(m[dR][curC] + " ");
                curC--;
            }
            while (curR != tR) {
                System.out.print(m[curR][tC] + " ");
                curR--;
            }
        }
    }

    public static void main(String[] args) {
        int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 },
                { 13, 14, 15, 16 } };
        spiralOrderPrint(matrix);

    }

}