蛇形矩阵

　　今天看到一个比较有趣的问题，输入一个整数n，输出一个n*n的矩阵，矩阵是逆时针方向增长的蛇形矩阵。这到底是个什么矩阵呢？举例来说:输入3，输出应该是

            1 2 3            8 9 4            7 6 5 

我的思路就是按照逆时针的方向来增大矩阵，num从1一直增大到n*n为止，增长方向由right,down,left,up4个bool值来控制，同时还设置了4个边界来控制i和j的终止范围，代码写得并不好，思路太过原始，变量也多，应该有一些简单的算法，不过暂时想把我这种思路记录下来吧，附代码:

#include<iostream>#include<iomanip>using namespace std;const int MAX = 10000;int a[MAX][MAX];//逆时针增加的蛇形矩阵 void snake(int N){    bool right = true, down = false, left = false, up = false;  //4个方向     int startOfI = 0, startOfJ = 0;    //4个边界     int boundOfJRight = N, boundOfJLeft = 0;    int boundOfIBottom = N, boundOfITop = 0;    //i和j的起始位置    int i = startOfI, j = startOfJ;    int num = 0;    while(num < N*N){        //先向右增加         if(right){            for(j = startOfJ; j < boundOfJRight; j++){                num++;                a[startOfI][j] = num;                cout << "第" << startOfI << "行\t第" <<j << "列" << num << endl;             }            right = !right;            down = !down;            startOfJ = j-1;            startOfI++;            //向右增加完成后，顶层肯定会铺满一层，所以顶层边界要加1            boundOfITop++;        }        //向下         if(down){            for(i = startOfI; i < boundOfIBottom; i++){                num++;                a[i][startOfJ] = num;                cout << "第" << i << "行\t第" << startOfJ << "列" << num << endl;            }            down = !down;            left = !left;            startOfI = i-1;            startOfJ--;            //同理，向下增加完成后，右边肯定会铺满一列，所以右边边界要减1            boundOfJRight--;        }        //向左         if(left){            for(j = startOfJ; j >= boundOfJLeft; j--){                num++;                a[startOfI][j] = num;                cout << "第" << startOfI << "行\t第" <<j << "列" << num << endl;            }            left = !left;            up = !up;            startOfJ = j+1;            startOfI--;            //向左增加完成后，底层肯定会铺满一层，所以底层边界要减1            boundOfIBottom--;        }        //向上         if(up){            for(i = startOfI; i >= boundOfITop; i--){                num++;                a[i][startOfJ] = num;                cout << "第" << i << "行\t第" << startOfJ << "列" << num << endl;            }            up = !up;            right = !right;            startOfI = i+1;            startOfJ++;            //向上增加完成后，左边肯定会铺满一列，所以左边边界要加1            boundOfJLeft++;        }    }    //打印     for(int k = 0; k < N; k++){        for(int p = 0; p < N; p++){            cout <<setw(5) << a[k][p] ;        }        cout << endl;    }}int main(){    int n;    cin >> n;    snake(n);}

运行结果示例：

n = 6