【左神算法】基础班第三课(二)——矩阵打印

顺时针顺序打印

记录圈数， 然后一圈一圈的打印。注意最后一圈只有一行或者一列的情况

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;


vector f(vector> v)
{
    vector res;
    if (v.size() != 0)
    {
        int level = (int)(min(v[0].size() - 1, v.size() - 1)) / 2 + 1;
        int length = static_cast(v.size()), width = (int)v[0].size();
        for (int i = 0; i < level; i++)
        {
            for (int j = i; j < width - i; j++)
            {
                res.push_back(v[i][j]);
            }
            
            for (int j = i + 1; j < length - i; j++)
            {
                res.push_back(v[j][width - i - 1]);
            }
            
            for (int j = width - i - 2; length - 2 * i != 1 && j >= i; j--)  // 仅剩一行不用管
            {
                res.push_back(v[length - i - 1][j]);
            }
            
            for (int j = length - i - 2; width - 2 * i != 1 && j > i; j--)  //仅剩一列不用管
            {
                res.push_back(v[j][i]);
            }
        }
    }
    return res;
}

int main()
{
    vector> v;
    int N = 5, M = 5;
    for (int i = 0; i < N; i++)
    {
        vector temp;
        for (int j = 0; j < M; j++)
        {
            temp.push_back(i * M + j);
        }
        v.push_back(temp);
    }
    
    for (int i = 0; i < N; i++)
    {
        vector temp;
        for (int j = 0; j < M; j++)
        {
            printf("%2d ", v[i][j]);
        }
        putchar(10);
    }
 
    auto res = f(v);
    cout << res.size() << endl;
    for (auto i : res)
    {
        cout << i << " ";
    }
    putchar(10);
    return 0;
}

z形打印， 利用副对角线i + j 属于[0, width + length - 2]的规律，按行遍历，控制列的范围即可。同时注意从上往下还是从下往上。 

 

vector f2(vector> v)
{
    vector res;
    if (v.size() != 0)
    {
        int length = (int)v.size();
        int width = (int)v[0].size();
        int cnt = length + width - 2;
        
        for (int i = 0; i <= cnt; i++)
        {
            if (i & 1)
            {
                for (int j = length - 1; j >= 0; j--)
                {
                    if (i - j >= 0 && i - j < width)
                    {
                        res.push_back(v[j][i - j]);
                    }
                }
            }
            else
            {
                for (int j = 0; j < length; j++)
                {
                    if (i - j >= 0 && i - j < width)
                    {
                        res.push_back(v[j][i - j]);
                    }
                }
            }
        }
    }
    
    
    return res;
}

 旋转90度，规律太简单，不写了。