我对带有指针输入的递归函数有疑问

Question

所以我正在尝试创建一个函数以螺旋顺序填充 1 个矩阵 有点像这样

1 2 3
8 9 4
7 6 5

我正在尝试先填充外部，然后使用相同类型的递归函数来填充不小的内部，但我只有 1 个内部：

1 2 3 4
12 1 1 5
11 1 1 6
10 9 8 7

这里是我使用的代码

#include <iostream>
#include <vector>
using namespace std;
void spiral_making(int sr,int sc,int er,int ec,int counter,vector< vector<int>>*so1){
    
    //sr :starting row
    //sc : stating column
    //er : ending row
    //ec : ending column
    int i; //iterator
    //insert the first row into the matrix
    if((sr-er==1)||(sc-ec==1)){(*so1)[sr][sc]= counter;}
    {
    for(i=sc;i<ec;i++){
        (*so1)[sr][i]=counter;
        counter++;  
    }
    
    //increse the starting row  by 1 
    sr++;
    //insert the back
    for (i=sr;i<er;i++){
        (*so1)[i][ec-1]=counter;
        counter++;
    }
    //then I reduced the ending column by 1
    //insert the bottom row (but it will be back ward) 

    ec--;
    for(i=ec;i>sc;i--){
        (*so1)[er-1][i-1]=counter;
        counter++;
    }
    //then i insert the left size column(but it also go backward)
    for(i=er-1;i>sr;i--){
        (*so1)[i-1][sc]=counter;
        counter++;
    }
    //then I icrease starting column and decrease ending row
    er--;
    sc++;
    //check if there are any small matrix inside if yes then conduct this function again with the smaller matrix
    if((er<sr)&&(sc<ec)){
        spiral_making(sr,sc,er,ec,counter,&(*so1));
    }
    }
}
int main(){
    
    //input section
    int n;
    do{
        std::cout<<"please input the  number greater than 1: ";
        cin>>n;
    }
    while(n<0);
    //try to work out with the matrix 
    vector< vector<int> > so(n,vector<int>(n,1));
    spiral_making(0,0,n,n,1,&so);
    //print the output
    for (int u=0;u<n;u++){
        for(int y=0;y<n;y++){
            std::cout<<so[u][y]<<" ";
        }
        std::cout<<std::endl;

    }
    return 0;
    

}

我有点新，所以请原谅我这个愚蠢的问题:)))

Answer 1

这是使用std::array 代替的一种方法。也许您可以根据需要对其进行调整。

template <size_t N>
using squareArr = std::array<std::array<int, N>, N>;

template <size_t N>
void spiral(squareArr<N>& mat) {
    if (mat.empty()) return;

    // Counter 
    size_t cnt{0};

    // Limits of the current circle
    size_t top{0}, left{0}, right{mat[0].size()-1}, bottom{mat.size()-1};

    while (top < bottom && left < right) {
        // Assign clockwise in the outer circle
        for (auto i=left,    j=top;    i<=right;  i++) mat.at(j).at(i) = cnt++;
        for (auto i=right,   j=top+1;  j<=bottom; j++) mat.at(j).at(i) = cnt++;
        for (auto i=right-1, j=bottom; i>left;    i--) mat.at(j).at(i) = cnt++;
        for (auto i=left,    j=bottom; j> top;    j--) mat.at(j).at(i) = cnt++;

        // Get a smaller circle
        top++, left++, right--, bottom--;
    }

    // Center (if N is odd)
    if (top == bottom && left == right) {
        mat[left][top] = cnt++;
    }
}

可测试here