【剑指offer】29.顺时针打印矩阵,旋转正方形矩阵

题目描述

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字，例如，如果输入如下4 X 4矩阵：

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

则依次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.

思路

public int[] spiralOrder(int[][] matrix) {
        if (matrix.length == 0) return new int[0];
        int rows = matrix.length;
        int columns = matrix[0].length;
        int[] res = new int[rows * columns];
        int index = 0;


        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {
            // 从左向右
            for (int i = left; i <= right; i++) res[index++] = matrix[top][i];
            // 从上向下
            for (int i = top + 1; i <= bottom; i++) res[index++] = matrix[i][right];
            // 从右向左,考虑重复：上下两行相等的时候
            if (top != bottom)
                for (int i = right - 1; i >= left; i--) res[index++] = matrix[bottom][i];
            // 从下向上，考虑重复：左右两行相等的时候
            if (left != right)
                for (int i = bottom - 1; i > top; i--) res[index++] = matrix[i][left];
            left++; right--; top++; bottom--;
        }
        return res;
    }

换个写法

public static void spiralOrderPrint(int[][] matrix) {
        // 左上角的行和列
        int top = 0;
        int left = 0;
        // 右下角的行和列
        int bottom = matrix.length - 1;
        int right = matrix[0].length - 1;

        while (left <= right && top <= bottom) {
            printEdge(matrix, left++, top++, right--, bottom--);
        }
    }

    private static void printEdge(int[][] matrix, int left, int top, int right, int bottom) {
        if (left == right) {
            for (int i = top; i <= bottom; i++) {
                System.out.println(matrix[i][left] + " ");
            }
        } else if (bottom == top) {
            for (int i = left; i <= right; i++) {
                System.out.println(matrix[top][i]);
            }
        } else {
            for (int i = left; i <= right; i++) System.out.print(matrix[top][i] + " ");
            // 从上向下
            for (int i = top + 1; i <= bottom; i++) System.out.print(matrix[i][right] + " ");
            // 从右向左
            for (int i = right - 1; i >= left; i--) System.out.print(matrix[bottom][i]+" ");
            // 从下向上
            for (int i = bottom - 1; i > top; i--) System.out.print(matrix[i][left]+ " ");
        }
    }


    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);
    }

旋转正方形矩阵

给定一个整型正方形矩阵matrix，请把矩阵调整成顺时针旋转90度的样子
【要求】额外空间复杂度为O(1)

public class RotateMatrix {
    public static void rotate(int[][] matrix) {

        int left = 0;
        int top = 0;
        int bottom = matrix.length - 1;
        int right = matrix[0].length - 1;
        while (left < right) {
            rotateEdge(matrix, left++, top++, bottom--, right--);
        }
    }

    private static void rotateEdge(int[][] matrix, int left, int top, int bottom, int right) {
        int times = right - left;
        int tmp = 0;
        for (int i = 0; i < times; i++) {
            tmp = matrix[top][left + i];
            matrix[top][left + i] = matrix[bottom - i][left];
            matrix[bottom - i][left] = matrix[bottom][right - i];
            matrix[bottom][right - i] = matrix[top + i][right];
            matrix[top + i][right] = tmp;
        }
    }

    private 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();
        }
    }

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