zigzag矩阵按照之字形线路产生方法(C++)
#include<iostream> using namespace std; void zigzag(int**M, int N) { /** example N=5 结果 * 0 1 5 6 14 15 27 28 * 2 4 7 13 16 26 29 42 * 3 8 12 17 25 30 41 43 * 9 11 18 24 31 40 44 53 * 10 19 23 32 39 45 52 54 * 20 22 33 38 46 51 55 60 * 21 34 37 47 50 56 59 61 * 35 36 48 49 57 58 62 63 **/ /*prestep表示上一步的动作,其值及其对应关系如下: case 1:上一步向右走 case 2;上一步左下走 case 3:上一步向下走 case 4:上一步右上走*/ int prestep; int count = 0; int i = 0; int j = 0; *(*(M + i) + j) = count; count++; prestep = 4; while (count < N*N) { if (i == 0)//上边界走法 { if (j == N - 1) { if (prestep == 1)//上一步右,接着左下 { i++; j--; prestep = 2; } else if (prestep == 4) { i++; prestep = 3; } } else if (prestep == 4)//上一步右上方向走,则现在向右走一步; { j++; prestep = 1; } else if (prestep == 1)//上一步向右走,现在需要向左下走一步; { i++; j--; prestep = 2; } else { cerr << "error 1!" << endl; return; } } else if (i == N - 1)//下边界走法 { if (prestep == 2)//上一步左下方向,则向右走 { j++; prestep = 1; } else if (prestep == 1 || prestep == 3)//上一步向右走,则像右上走; { i--; j++; prestep = 4; } else { cerr << "error 2!" << endl; return; } } else if (j == 0)//左边界 { if (prestep == 2)//上一步左下,这一步向下 { i++; prestep = 3; } else if (prestep == 3)//上一步向下,这一步右上 { i--; j++; prestep = 4; } else { cerr << "error 3!" << endl; return; } } else if (j == N - 1)//右边界 { if (prestep == 4)//上一步右上,这一步向下 { i++; prestep = 3; } else if (prestep == 3 || prestep == 1)//上一步向下,这一步左下 { i++; j--; prestep = 2; } else { cerr << "error 4!" << endl; return; } } else { if (prestep == 2)//上一步左下,接着左下 { i++; j--; prestep = 2; } else if (prestep == 4)//上一步右上,这一步接着右上 { i--; j++; prestep = 4; } else { cerr << "error 5!" << endl; return; } } *(*(M + i) + j) = count++; } } int main() { int N; cout << "please input N:"; cin >> N; int **M = (int**)malloc(sizeof(int *)*N); for (int i = 0; i < N; i++) { *(M + i) = (int *)malloc(sizeof(int)*N); } zigzag(M, N); for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { cout << *(*(M+i)+j) << "\t"; } cout << endl; } for (int i = 0; i < N; i++) { free(*(M + i)); } free(M); }