Question

我正在编写一个代码来对2D矩阵执行递归螺旋扫描。当行数和列数相同时，它工作正常，但不是这样（当num rows＆gt; num列，反之亦然）。螺旋扫描以螺旋顺序打印矩阵。例如：

输入矩阵：
  1 2 3
  4 5 6
  7 8 9
  输出：
  1 2 3 6 9 8 7 4 5

请帮助查找错误。这是我的代码：

#include <stdio.h>
#include <stdlib.h>
int a[10][10];
void spiral_scan(int a[][10],int sr,int sc,int er,int ec)
{
    if((ec<1)||(er<1))
      return ;
    int i;
    for(i=sr;i<=ec;i++)
        printf("%d ",a[sr][i]);
    //printf("\n");
    sr++;
    for(i=sr;i<=er;i++)
        printf("%d ",a[i][ec]);
    ec--;
    for(i=ec;i>=sc;i--)
        printf("%d ",a[er][i]);
   er--;
    for(i=er;i>=sr;i--)
        printf("%d ",a[i][sc]);

    sc++;
    spiral_scan(a,sr,sc,er,ec);
}

int main()
{
    printf("\t\tSPIRAL SCAN OF MATRIX RECURSUVELY\n");
    scanf("Enter the number of rows and coloumns:\n");
    int m1,n1,i,j;
    scanf("%d%d",&m1,&n1);

    printf("Enter the elements in the matrix:\n");
    for(i=0;i<m1;i++)
    {
        for(j=0;j<n1;j++)
            scanf("%d",&a[i][j]);
    }
    printf("The entered matrix is:\n");
    for(i=0;i<m1;i++)
    {
        printf("\n");
        for(j=0;j<n1;j++)
            printf("%d ",a[i][j]);
    }
    printf("\n");
    spiral_scan(a,0,0,m1-1,n1-1);
    return 0;
}

Answer 1

这是我在C ++中的代码。

//to check for boundary conditions
bool check(int i,int j,int rows,int cols)
{
    if(i>=0 && j>=0 && i<rows && j<cols)
        return 1;
    return 0;
}

//to check if all elements have been traversed
bool brk(int rows,int cols,vector<vector<int>> mark)
{
    for(int i=0;i<rows;i++)
    {
        for(int j=0;j<cols;j++)
        {
            if(mark[i][j]==0)
                return 0;
        }
    }
    return 1;
}

vector<int> spiralOrder(vector<vector<int>>& matrix) {

    vector < int > V;
    if(matrix.size()==0)
        return V;

    int rows = matrix.size();
    int cols = matrix[0].size();

    vector < vector<int> > mark(rows , vector<int>(cols,0) );

    int i=0,j=0;

    while(1)
    {
        //when going right
        while(check(i,j,rows,cols) && mark[i][j]==0)
        {
            mark[i][j]=1;
            V.push_back(matrix[i][j]);
            j++;
        }
        j--;    i++;
        if(brk(rows,cols,mark)==1)
            break;

        //when going down
        while(check(i,j,rows,cols) && mark[i][j]==0)
        {
            mark[i][j]=1;
            V.push_back(matrix[i][j]);
            i++;
        }
        i--;    j--;
        if(brk(rows,cols,mark)==1)
            break;

        //when going left
        while(check(i,j,rows,cols) && mark[i][j]==0)
        {
            mark[i][j]=1;
            V.push_back(matrix[i][j]);
            j--;
        }
        j++;    i--;
        if(brk(rows,cols,mark)==1)
            break;

        //when going up
        while(check(i,j,rows,cols) && mark[i][j]==0)
        {
            mark[i][j]=1;
            V.push_back(matrix[i][j]);
            i--;
        }
        i++;    j++;
        if(brk(rows,cols,mark)==1)
            break;
    }
    return V;
}

这是我在Leetcode中接受的代码。问题链接 - https://leetcode.com/problems/spiral-matrix/

递归解决方案。

class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix)
{
    vector<int> ret;
    if (!matrix.size())return ret;
    int direc = 1;
    set<pair<int, int>> exist;
    pair<int, int> start{ 0, 0 };
    direction(matrix, direc, exist, start, ret);
    return ret;

}
void direction(vector<vector<int>>& matrix, int &direc, set<pair<int, int>> &exist, pair<int, int> &start, vector<int>& ret)
{
    if (start.first>int(matrix.size()) - 1 || start.second>int(matrix[0].size()) - 1 || exist.count(start))
        return;
    int j = 0;
    switch (direc)
    {
    case 1: //east

        for (j = start.second; j<matrix[0].size() && !exist.count(make_pair(start.first, j)); j++)
        {
            ret.push_back(matrix[start.first][j]);
            exist.insert(make_pair(start.first, j));
        }
        direc = 2;
        start = make_pair(start.first + 1, j - 1);
        direction(matrix, direc, exist, start, ret);
        break;
    case 2://south

        for (j = start.first; !exist.count(make_pair(j, start.second)) && j<matrix.size(); j++)
        {
            ret.push_back(matrix[j][start.second]);
            exist.insert(make_pair(j, start.second));
        }
        direc = 3;
        start = make_pair(j - 1, start.second - 1);
        direction(matrix, direc, exist, start, ret);
        break;


    case 3://west

        for (j = start.second; !exist.count(make_pair(start.first, j)) && j >= 0; j--)
        {
            ret.push_back(matrix[start.first][j]);
            exist.insert(make_pair(start.first, j));
        }
        direc = 4;
        start = make_pair( start.first - 1,j+1);
        direction(matrix, direc, exist, start, ret);
        break;

    case 4://north

        for (j = start.first; !exist.count(make_pair(j, start.second)) && j >= 0; j--)
        {
            ret.push_back(matrix[j][start.second]);
            exist.insert(make_pair(j, start.second));
        }
        direc = 1;
        start = make_pair(j + 1, start.second + 1);
        direction(matrix, direc, exist, start, ret);
        break;
    }
}
};

矩阵的递归螺旋扫描

1 个答案: