顺时针打印矩阵

发布时间 2023-03-27 00:36:34作者: zhengbiyu

 

 

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if(matrix.length == 0) return new int[0];
        int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0;
        int[] res = new int[(r + 1) * (b + 1)];
        while(true) {
            //从左往右
            //列在变,列为循环值
            //从左往右的下一步是往下走,上边界内缩,故++t
            for(int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right.
            if(++t > b) break;
            //从上往下,行在变
            //从上往下的下一步是从右往左,右边界收缩,--r
            for(int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom.
            if(l > --r) break;
            //从右向左,列在变
            //从右往左的下一步是从下往上,下边界收缩,--b
            for(int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left.
            if(t > --b) break;
            //从下到上,行在变
            //从下到上的下一步是从左到右,左边界收缩,++l
            for(int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top.
            if(++l > r) break;
        }
        return res;
    }
}