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顺时针打印矩阵

顺时针打印矩阵

作者: 怎样会更好 | 来源:发表于2018-10-31 19:01 被阅读0次

    题目:

    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.

    public static ArrayList<Integer> printMatrix(int[][] matrix) {
        if (matrix.length == 0) {
            return null;
        }
        ArrayList<Integer> list = new ArrayList<>();
        if (matrix.length == 1) {
            for (int i = 0; i < matrix[0].length; i++) {
                list.add(matrix[0][i]);
            }
            return list;
        }
        if (matrix[0].length == 1) {
            for (int i = 0; i < matrix.length; i++) {
                list.add(matrix[i][0]);
            }
            return list;
        }
        int[][] map = new int[matrix.length + 2][matrix[0].length + 2];
        //初始化地图
        for (int i = 0; i < map.length; i++) {
            for (int j = 0; j < map[0].length; j++) {
                if (i == 0 || j == 0 || i == map.length - 1 || j == map[0].length - 1) {
                    map[i][j] = 1;
                } else {
                    map[i][j] = 0;
                }
            }
        }
        for (int i = 1; i < map.length - 1; ) {
            for (int j = 1; j < map[0].length - 1; ) {
                if (map[i][j] == 0) {
                    list.add(matrix[i - 1][j - 1]);
                    map[i][j] = 1;
                }
                //如果右边有
                if (map[i][j + 1] == 0 && map[i - 1][j] == 1) {
                    j++;
                    continue;
                }
                //下边有
                if (map[i + 1][j] == 0) {
                    i++;
                    continue;
                }
                //左边
                if (map[i][j - 1] == 0) {
                    j--;
                    continue;
                }
                if (map[i - 1][j] == 0) {
                    i--;
                    continue;
                }
                if (map[i + 1][j] == 1 && map[i][j + 1] == 1 && map[i - 1][j] == 1 && map[i][j - 1] == 1) {
                    break;
                }
    
            }
            break;
        }
        return list;
    }
    
    

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