54. Spiral Matrix

Given a matrix ofmxnelements (mrows,ncolumns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

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

You should return[1,2,3,6,9,8,7,4,5].

S:

从四周到中间分行列存入数组

    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int n = matrix.size();
        if (n == 0) return vector<int>{};
        int m = matrix[0].size();
        vector<int> result(m * n);
        int maxLevel = ceil(min(m, n) / 2.0);
        int idx = 0;
        for (int i = 0; i < maxLevel; ++i) {
            int row1 = i, row2 = n - i - 1, col1 = i, col2 = m - i - 1;
            for (int j = col1; j <= col2; ++j) {
                result[idx++] = matrix[row1][j];
            }
            for (int j = row1 + 1; j <= row2; ++j) {
                result[idx++] = matrix[j][col2];
            }
            if (row2 > row1) {
                for (int j = col2 - 1; j >= col1; --j) {
                    result[idx++] = matrix[row2][j];
                }
            }
            if (col2 > col1) {
                for (int j = row2 - 1; j > row1; --j) {
                    result[idx++] = matrix[j][col1];
                }
            }
        }
        return result;
    }

498. Diagonal Traverse

Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.

Example:

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

Explanation:

Note:

1. The total number of elements of the given matrix will not exceed 10,000.

S

    vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
        int m = matrix.size();
        if (m == 0) return vector<int>{};
        int n = matrix[0].size();
        vector<int> result(m * n);
        vector<int> dx{-1, 1};
        vector<int> dy{1, -1};
        int dir = 0;
        for (int x = 0, y = 0, i = 0; i < m * n; ++i) {
            result[i] = matrix[x][y];
            int idx = dir % 2;
            if (y < n - 1 && ((x == 0 && idx == 0) || (x == m - 1 && idx == 1))) {
                y++;
                dir++;
            } else if (x < m - 1 && ((y == 0 && idx == 1) || (y == n - 1 && idx == 0))) {
                x++;
                dir++;
            } else {
                x += dx[idx];
                y += dy[idx];
            }
        }
        return result;
    }