54. Spiral Matrix

This is an exactly same problem with "59. Spiral Matrix II". I set every direction as a status, when one direction was implemented, the current status turns to the next status.

Which different status, the scan direction is different, but all scans cannot exceed the edges of the matrix.

The time complexity is O(m*n):

    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<>();
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
            return res;
        int m = matrix.length;
        int n = matrix[0].length;
        int top = 0, bottom = m - 1, left = 0, right = n - 1;

        int status = 0;  //we set every directions as a status, there are toally 4 status.
        while (top <= bottom && left <= right) {
            if (status % 4 == 0) {  //the first direction, from left to right
                for (int j = left; j <= right; j++) {
                    res.add(matrix[top][j]);
                }
                top++;

            } else if (status % 4 == 1) {  //the second direction, from top to bottom
                for (int i = top; i <= bottom; i++) {
                    res.add(matrix[i][right]);
                }
                right--;
            } else if (status % 4 == 2) {  //the third direction, from right to left
                for (int j = right; j >= left; j--) {
                    res.add(matrix[bottom][j]);
                }
                bottom--;
            } else if (status % 4 == 3) {  //the fourth direction, from bottom to top
                for (int i = bottom; i >= top; i--) {
                    res.add(matrix[i][left]);
                }
                left++;
            }
        }
        return res;
    }

标签：status,matrix,bottom,int,54,top,Spiral,res,Matrix
来源： https://www.cnblogs.com/feiflytech/p/15862380.html