思路:遍历框框
左上角 和右下角的点,可以确定一个矩阵。
左上角(tr,tc)
右下角(dr,dc);
public static void PrintMatrix(int [][] matrix){ int tr=0; int tc=0; int dr=matrix.length-1; int dc=matrix[0].length-1; while(tc<=dc&&tr<=dr){
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--; } }
旋转
public static void rotateEdge(int[][] m, int tR, int tC, int dR, int dC) { int times = dC - tC; int tmp = 0; for (int i = 0; i != times; i++) { tmp = m[tR][tC + i]; m[tR][tC + i] = m[dR - i][tC]; m[dR - i][tC] = m[dR][dC - i]; m[dR][dC - i] = m[tR + i][dC]; m[tR + i][dC] = tmp; } } public static void printMatrix(int[][] matrix) { for (int i = 0; i != matrix.length; i++) { for (int j = 0; j != matrix[0].length; j++) { System.out.print(matrix[i][j] + " "); } System.out.println(); } }
顺时针打印矩阵/90度旋转矩阵
原文地址:https://www.cnblogs.com/bowenqianngzhibushiwo/p/11629321.html
时间: 2024-11-09 04:52:18