leetcode: Spiral Matrix

原创于 2013-05-28 00:47:01 发布 · 651 阅读

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本文介绍了一种矩阵螺旋遍历算法的实现方法，通过C++代码详细解析了如何按螺旋顺序遍历一个m x n的二维矩阵，并返回所有元素。

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Given a matrix of m x n elements (m rows, n columns), 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].

class Solution {

    int min(int a, int b)
    {
        if (a > b)
            return b;
        return a;
    }
public:
    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        vector<int> retVector;
        
        if (matrix.size() == 0)
            return retVector;
            
        int rows = matrix.size();
        int cols = matrix[0].size();
        int maxLoop = min(rows, cols);
        
        
        for (int loop = 0; loop < maxLoop/2; loop++)
        {
            for (int i = loop; i<cols-loop; i++)
                retVector.push_back(matrix[loop][i]);
            
            for (int j = loop+1; j<rows-loop; j++)
                retVector.push_back(matrix[j][cols-1-loop]);
                    
            for (int k = cols-1-loop-1; k>=loop; k--)
                retVector.push_back(matrix[rows-1-loop][k]);
            
            for (int l = rows-1-loop-1; l>=loop+1; l--)
                retVector.push_back(matrix[l][loop]);
        }
        
        if (maxLoop%2)
        {
            if (rows < cols)
            {
                for (int i=maxLoop/2; i<cols-maxLoop/2; i++)
                    retVector.push_back(matrix[maxLoop/2][i]);
            }
            else if (rows > cols)
            {
                for (int j=maxLoop/2; j<rows-maxLoop/2; j++)
                    retVector.push_back(matrix[j][maxLoop/2]);
            }
            else
            {
                retVector.push_back(matrix[maxLoop/2][maxLoop/2]);
            }
        }
        return retVector;
    }
};