順時針打印矩陣

 

 

 

 

public class Code_06_PrintMatrixSpiralOrder {

	public static void spiralOrderPrint(int[][] matrix) {
		int tR = 0;  //左上角的行
		int tC = 0;  //左上角的列
		int dR = matrix.length - 1;      //右上角的行
		int dC = matrix[0].length - 1;   //右下角的列
		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) {                          
			for (int i = tC; i <= dC; i++) {                        
				System.out.print(m[tR][i] + " ");
			}
		} else if (tC == dC) {
			for (int i = tR; i <= dR; i++) {
				System.out.print(m[i][tC] + " ");
			}
		} else {
			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);

	}

}

輸出：

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10

思路蠻簡單：

這一段代碼，是包含了兩個邊界狀況，當二維數組退化爲一位數組時，應該怎麼處理。下面纔是開始的邏輯。