leetcode 54|59. Spiral Matrix 1|2

54. Spiral Matrix

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].

旋转读取二维矩阵，x,y是每一圈的左上角（定点坐标）的横纵坐标，row是每一圈的行数，col是每一圈的列数。

每一圈扫完，定点坐标往右下移动，row-2, col-2

class Solution {
public:
    void help(vector<vector<int>>& matrix, vector<int> &ret, int x, int y, int row, int col)
    {
        if (row == 0 || col == 0)    
            return;
        else if (col == 1)
        {
            for (int i = x; i < x + row; i++)
                ret.push_back(matrix[i][y]);
            return;
        }
        else if (row == 1)
        {
            for (int i = y; i < y + col; i++)
                ret.push_back(matrix[x][i]);
            return;
        }
        else
        {
            for (int i = y; i < y + col; i++)  
                ret.push_back(matrix[x][i]);
            for (int i = x + 1; i < x + row - 1; i++)
                ret.push_back(matrix[i][y + col - 1]);
            for (int i = y + col - 1; i >= y; i--)  
                ret.push_back(matrix[x + row - 1][i]);
            for (int i = x + row - 2; i > x; i--)
                ret.push_back(matrix[i][y]);
        
            help(matrix, ret, x + 1, y + 1, row - 2, col - 2);
        }
        return;
    }
    
    vector<int> spiralOrder(vector<vector<int>>& matrix) 
    {
        vector<int> ret;
        if (matrix.size() == 0)  return ret;
        int row = matrix.size();
        int col = matrix[0].size();
        help(matrix, ret, 0, 0, row, col);
        return ret;
    }
};

59. Spiral Matrix II

Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n = 3,

You should return the following matrix:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

way-1:找规律输出就行

class Solution {
public:
    void Matrix(int n, int pos, int count, vector<vector<int>> &result)
    {
        //上面一行输入
        for (int i = pos; i < n - pos; i++)
            result[pos][i] = count ++;

        
        //右边一列
        for(int i = pos + 1; i < n - pos - 1; i++)
            result[i][n-1-pos] = count ++;

        
        if (count > n * n)  //上面和右边填入之后需要判断，因为可能只有一个坑，而后面可能会重新覆盖
            return;
        
        //下面一行
        for(int i = n - pos - 1; i >= pos; i--)
            result[n-1-pos][i] = count++;

        
        //左边一列
        for (int i = n - pos - 2; i > pos; i--)
            result[i][pos] = count++;
        
        if (count <= n * n)
           Matrix(n, pos+1, count, result);
        return;
    }
    
    vector<vector<int>> generateMatrix(int n) 
    {  
        vector<vector<int>> result;
        if (n <= 0)
            return result;
        else if (n == 1)
        {
            result.push_back(vector<int> (1,1));
            return result;
        }
        else
        {
            for(int i=0;i<n;i++)
            {
                result.push_back(vector<int> (n, 0));
            }
            Matrix(n, 0, 1, result);
        }
        return result;        
    }
};