[LeetCode]Spiral Matrix

最新推荐文章于 2019-07-07 11:13:21 发布

原创最新推荐文章于 2019-07-07 11:13:21 发布 · 4.5k 阅读

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本文提供了一种螺旋遍历二维矩阵的方法，通过四个方向的循环迭代实现从外向内逐层遍历矩阵元素。该算法适用于多种编程场景，如数据处理、图像处理等。

class Solution {
//key observe: when any component is done, its beginX,endX or beginY,endX will change
public:
	vector<int> spiralOrder(vector<vector<int> >& matrix) {
		vector<int> result;
		if (matrix.empty()) return result;
		int beginX = 0, endX = matrix[0].size() - 1;
		int beginY = 0, endY = matrix.size() - 1;
		while (true) {
			// From left to right
			for (int i = beginX; i <= endX; ++i)
				result.push_back(matrix[beginY][i]);
			if (++beginY > endY) break;
			// From top down
			for (int i = beginY; i <= endY; ++i)
				result.push_back(matrix[i][endX]);
			if (beginX > --endX) break;
			// From right to left
			for (int i = endX; i >= beginX; --i)
				result.push_back(matrix[endY][i]);
			if (beginY > --endY) break;
			// From bottom up
			for (int i = endY; i >= beginY; --i)
				result.push_back(matrix[i][beginX]);
			if (++beginX > endX) break;
		}
		return result;
	}
};

second time

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(matrix.size() == 0) return vector<int>();
        int x1 = 0;
        int y1 = 0;
        int x2 = matrix.size()-1;
        int y2 = matrix[0].size()-1;
        vector<int> ans;
        while(x1 <= x2 && y1 <= y2)
        {
            //up row
            for(int i = y1; i <= y2; ++i) ans.push_back(matrix[x1][i]);
            //right column
            for(int i = x1+1; i <= x2; ++i) ans.push_back(matrix[i][y2]);
            //bottom row
            if(x2 != x1)
            for(int i = y2-1; i >= y1; --i) ans.push_back(matrix[x2][i]);
            //left column
            if(y1 != y2)
            for(int i = x2-1; i > x1; --i) ans.push_back(matrix[i][y1]);
            
            x1++, y1++, x2--, y2--;
        }

        return ans;
    }
};