2014-03-18 01:45
题目:给定一个NxN的矩阵,就地旋转90度。(没有样例又不说方向的话,随便往哪儿转。)
解法:如果N为奇数,除了中心点以外四等分。如果N为偶数,四等分。按照A->B->C->D->A的方式,轮换赋值,需要O(1)的额外空间保存A的值。
代码:
1 // 1.6 Given an image represented by an NXN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place? 2 #include <cstdio> 3 #include <vector> 4 using namespace std; 5 6 class Solution { 7 public: 8 void rotateImage90(vector<vector<int> > &board) { 9 int n = (int)board.size(); 10 int i, j; 11 12 if (n < 2) { 13 return; 14 } 15 int w, h; 16 int tmp; 17 18 w = n / 2; 19 h = n - w; 20 for (i = 0; i < w; ++i) { 21 for (j = 0; j < h; ++j) { 22 tmp = board[i][j]; 23 board[i][j] = board[n - 1 - j][i]; 24 board[n - 1 - j][i] = board[n - 1 - i][n - 1 - j]; 25 board[n - 1 - i][n - 1 - j] = board[j][n - 1 - i]; 26 board[j][n - 1 - i] = tmp; 27 } 28 } 29 } 30 }; 31 32 int main() 33 { 34 vector<vector<int> > board; 35 int i, j, n; 36 Solution sol; 37 38 while (scanf("%d", &n) == 1 && n > 0) { 39 board.resize(n); 40 for (i = 0; i < n; ++i) { 41 board[i].resize(n); 42 } 43 44 for (i = 0; i < n; ++i) { 45 for (j = 0; j < n; ++j) { 46 scanf("%d", &board[i][j]); 47 } 48 } 49 sol.rotateImage90(board); 50 for (i = 0; i < n; ++i) { 51 printf("%d", board[i][0]); 52 for (j = 1; j < n; ++j) { 53 printf(" %d", board[i][j]); 54 } 55 printf(" "); 56 } 57 printf(" "); 58 59 for (i = 0; i < n; ++i) { 60 board[i].clear(); 61 } 62 board.clear(); 63 } 64 65 return 0; 66 }