剑指 Offer 29. 顺时针打印矩阵

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        int m = matrix.length;
        if(m == 0) return new int[]{};
        int n = matrix[0].length;
        if(m == 1) return matrix[0];
        if(n == 0) return new int[]{};
        int[] res = new int[m*n];
        int k = 0;
        int row_start=0,row_end = m,col_start = 0,col_end = n;
        while (row_start<row_end && col_start<col_end) {
            for (int i = col_start; i < col_end; i++) {
                res[k++] = matrix[row_start][i];
            }
            if(k == m*n) return res;
            row_start++;
            for (int j = row_start; j < row_end; j++) {
                res[k++] = matrix[j][col_end-1];
            }
            if(k == m*n) return res;
            col_end--;
            for (int i = col_end - 1; i >= col_start; i--) {
                res[k++] = matrix[row_end-1][i];
            }
            if(k == m*n) return res;
            row_end--;
            for (int j = row_end - 1; j >= row_start; j--) {
                res[k++] = matrix[j][col_start];
            }
            if(k == m*n) return res;
            col_start++;
        }
        return res;
    }
}