【leetcode】54. Spiral Matrix(螺旋矩阵

1、题目描述

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
  [ 1, 2, 3 ],
  [ 4, 5, 6 ],
  [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

2、问题描述：

• 螺旋的打印矩阵。

3、问题关键：

通用的经典做法：
1.定义方向：dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, 1, 0};
2.坐标移动：int a = x + dx[d],  b = y + dy[d];
3.边界条件：
if (a  < 0 || a == n || b < 0 || b == m || st[a][b])  {
     d = (d + 1) % 4;
     a = x + dx[d], b = y + dy[d];
}

4、C++代码：

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if (matrix.empty()) return vector<int> ();
        int n = matrix.size(), m = matrix[0].size();
        vector<vector<bool>> f(n, vector<bool>(m, false));
        int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};//向右y + 1，向下x + 1， 向左y — 1，向上x - 1.
        vector<int> res;
        int x = 0, y = 0, d = 0;
        for (int i = 0; i < m * n; i ++) {
            int a = x + dx[d], b = y + dy[d];
            if (a < 0 || a >= n || b < 0 || b >= m || f[a][b]){
                d = (d + 1) % 4;
                a = x + dx[d], b = y + dy[d];
            }
            res.push_back(matrix[x][y]);
            f[x][y] = true;
            x = a, y = b;
        }
        return res;
    }
};