[leetcode]Spiral Matrix

简单题。但做得不简洁。里面用了一个count来计数，还是不错的。因为有可能只有一行（或子状态中），那么m_low == m_high，这时，当m_low++后，m_high大于m_low，可能会从左往右和回扫都是同一行。

public class Solution {
    public ArrayList<Integer> spiralOrder(int[][] matrix) {
        ArrayList<Integer> ans = new ArrayList<Integer>();
        int m = matrix.length;
        if (m == 0) return ans;
        int n = matrix[0].length;
        if (n == 0) return ans;
        int m_low = 0;
        int m_high = m-1;
        int n_low = 0;
        int n_high = n-1;
        int count = m*n;
        while (count > 0) {
            for (int i = n_low; i <= n_high && count > 0; i++) {
                ans.add(matrix[m_low][i]);
                count--;
            }
            m_low++;
            for (int i = m_low; i <= m_high && count > 0; i++) {
                ans.add(matrix[i][n_high]);
                count--;
            }
            n_high--;
            for (int i = n_high; i >= n_low && count > 0; i--) {
                ans.add(matrix[m_high][i]);
                count--;
            }
            m_high--;
            for (int i = m_high; i >= m_low && count > 0; i--) {
                ans.add(matrix[i][n_low]);
                count--;
            }
            n_low++;
        }
        return ans;
    }
}

Python3

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m = len(matrix)
        if m == 0:
            return []
        n = len(matrix[0])
        if n == 0:
            return []
        ret = []
        i = j = 0
        cnt = 0
        direction = 0 # 0, 1, 2, 3
        upper_bound = 0
        lower_bound = m - 1
        left_bound = 0
        right_bound = n - 1
        
        upper_bound += 1
        while cnt < m * n:
            ret.append(matrix[i][j])
            if direction == 0:
                if j != right_bound:
                    j += 1
                else:
                    right_bound -= 1
                    i += 1
                    direction = 1
            elif direction == 1:
                if i != lower_bound:
                    i += 1
                else:
                    lower_bound -= 1
                    j -= 1
                    direction = 2
            elif direction == 2:
                if j != left_bound:
                    j -= 1
                else:
                    left_bound += 1
                    i -= 1
                    direction = 3
            elif direction == 3:
                if i != upper_bound:
                    i -= 1
                else:
                    upper_bound += 1
                    j += 1
                    direction = 0
            cnt += 1
        return ret