#374 Spiral Matrix

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本文介绍了一种将矩阵元素按螺旋顺序遍历的方法，并提供了一个通过边界控制实现螺旋遍历的具体算法实例。该算法适用于m x n大小的矩阵。

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.

Have you met this question in a real interview?

Yes

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].

题目思路：

这题和#381一样，只要设好了boundary，再注意下while loop break的条件，再仔细些就不会有问题。

Mycode（AC = 27ms）：

class Solution {
public:
    /**
     * @param matrix a matrix of m x n elements
     * @return an integer array
     */
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        // Write your code here
        vector<int> ans;
        
        int num_rows = matrix.size();
        if (num_rows == 0) return ans;
        
        ans.resize(num_rows * matrix[0].size());
        
        int upper_row = 0,
            down_row = num_rows - 1,
            left_col = 0,
            right_col = matrix[0].size() - 1,
            count = 0;
            
        while (count < ans.size()) {
            // get upper row
            for (int i = left_col; i <= right_col; i++) {
                ans[count++] = matrix[upper_row][i];
            }
            upper_row++;
            
            if (count >= ans.size()) break;
            
            // get right col
            for (int i = upper_row; i <= down_row; i++) {
                ans[count++] = matrix[i][right_col];
            }
            right_col--;
            
            if (count >= ans.size()) break;
            
            // get bottom row
            for (int i = right_col; i >= left_col; i--) {
                ans[count++] = matrix[down_row][i];
            }
            down_row--;
            
            if (count >= ans.size()) break;
            
            // get left col
            for (int i = down_row; i >= upper_row; i--) {
                ans[count++] = matrix[i][left_col];
            }
            left_col++;
        }
        
        return ans;
    }
};