54. Spiral Matrix

最新推荐文章于 2024-01-06 15:12:04 发布

原创最新推荐文章于 2024-01-06 15:12:04 发布 · 128 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种矩阵螺旋遍历的算法实现，通过定义四个方向的递归函数完成遍历，并利用二维布尔数组记录已访问元素，适用于m×n大小的矩阵。

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].

这道题，我们可以使用递归的方法来解决，分别定义四个函数right,down,left,up。

用一个二维vector<vector<bool>>来记录是否已经读取过这个数字了。

具体代码如下

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int>fuck;
        if(matrix.empty())return fuck;
	vector<vector<bool>>shit(matrix.size(), vector<bool>(matrix[0].size(), true));
	int m = (int)matrix[0].size();
	int n = (int)matrix.size();
	if(m==1)
	{
	    for(int i=0;i<n;i++)fuck.push_back(matrix[i][0]);
	    return fuck;
	}
	if(n==1)
	{
	    for(auto a:matrix[0])fuck.push_back(a);
	    return fuck;
	}
	int bitch = 0;
	right(bitch, fuck, matrix, m, n,shit);
	return fuck;
    }
void right(int bitch, vector<int> &fuck, vector<vector<int>>matrix, int m, int n, vector<vector<bool>>shit)
{
	for (int i = bitch; i < m - bitch; i++)
	{
		fuck.push_back(matrix[bitch][i]);
		shit[bitch][i] = false;
	}
	if (shit[bitch+1][m - bitch-1])
	{
		bitch++;
		down(bitch, fuck, matrix, m, n, shit);
	}
}
void down(int bitch, vector<int>&fuck, vector<vector<int>>matrix, int m, int n, vector<vector<bool>>shit)
{
	for (int i = bitch; i < n - bitch+1 ; i++)
	{
		fuck.push_back(matrix[i][m - bitch]);
		shit[i][m - bitch] = false;
	}
	if (shit[n - bitch][m - bitch-1])
	{
		left(bitch, fuck, matrix, m, n, shit);
	}
}
void left(int bitch, vector<int>&fuck, vector<vector<int>>matrix, int m, int n, vector<vector<bool>>shit)
{
	for (int i = m - bitch - 1; i >= bitch - 1; i--)
	{
		fuck.push_back(matrix[n - bitch][i]);
		shit[n - bitch][i] = false;
	}
	if (shit[n - bitch - 1][bitch - 1])
	{
		up(bitch, fuck, matrix, m, n, shit);
	}
}
void up(int bitch, vector<int>&fuck, vector<vector<int>>matrix, int m, int n, vector<vector<bool>>shit)
{
	for (int i = n - 1 - bitch; i >= bitch; i--)
	{
		fuck.push_back(matrix[i][bitch - 1]);
		shit[i][bitch - 1] = false;
	}
	if (shit[bitch][bitch])
	{
		right(bitch, fuck, matrix, m, n, shit);
	}
}
};

请别太在意代码里面的fuck,shit以及bitch，毕竟他们是万能的~