LeetCode54 Spiral Matrix 螺旋矩阵

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]

题源：here；完整实现：here
思路：
剥洋葱。我们像剥洋葱一样从外往内一层一层的输出。两种实现方案：1 单变量版；2 4变量版。4变量版的简洁和可读性远超单变量版。
实现1–单变量版

    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> result;
        if (matrix.size() == 0) return result;
        int rows = matrix.size(), cols = matrix[0].size();
        if (rows == 1){
            for (int i = 0; i < cols; i++) result.push_back(matrix[0][i]);
            return result;
        }
        if (cols == 1){
            for (int i = 0; i < rows; i++) result.push_back(matrix[i][0]);
            return result;
        }

        for (int i = 0; i < min(rows, cols) / 2; i++){
            for (int j = i; j < cols - i - 1; j++){
                result.push_back(matrix[i][j]);
            }
            for (int j = i; j < rows - i - 1; j++){
                result.push_back(matrix[j][cols-i-1]);
            }
            for (int j = i; j < cols - i - 1; j++){
                result.push_back(matrix[rows-i-1][cols - j - 1]);
            }
            for (int j = i; j < rows - i - 1; j++){
                result.push_back(matrix[rows - j - 1][i]);
            }
        }

        if (cols >= rows && rows % 2){
            for (int i = rows / 2; i < cols - rows/2; i++){
                result.push_back(matrix[rows/2][i]);
            }
        }

        else if (cols < rows && cols % 2){
            for (int i = cols / 2; i < rows - cols/2; i++){
                result.push_back(matrix[i][cols/2]);
            }
        }
        return result;
    }

实现2–4变量版

    vector<int> spiralOrder_2(vector<vector<int>>& matrix){
        vector<int> result; if (matrix.size() == 0) return result;
        int left = 0, right = matrix[0].size() - 1, top = 0, bottom = matrix.size() - 1;
        while (left <= right && top <= bottom){
            for (int i = left; i < right; i++) result.push_back(matrix[top][i]);
            for (int i = top; i <= bottom; i++) result.push_back(matrix[i][right]);
            if (left < right && top < bottom){
                for (int i = right - 1; i > left; i--) result.push_back(matrix[bottom][i]);
                for (int i = bottom; i > top; i--) result.push_back(matrix[i][left]);
            }
            left++, right--, top++, bottom--;
        }

        return result;
    }