《Cracking the Coding Interview》——第1章：数组和字符串——题目6

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：

 // 1.6 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?
 #include <cstdio>
 #include <vector>
 using namespace std;

 class Solution {
 public:
     void rotateImage90(vector<vector<int> > &board) {
         int n = (int)board.size();
         int i, j;

         if (n < ) {
             return;
         }
         int w, h;
         int tmp;

         w = n / ;
         h = n - w;
         for (i  = ; i < w; ++i) {
             for (j = ; j < h; ++j) {
                 tmp = board[i][j];
                 board[i][j] = board[n -  - j][i];
                 board[n -  - j][i] = board[n -  - i][n -  - j];
                 board[n -  - i][n -  - j] = board[j][n -  - i];
                 board[j][n -  - i] = tmp;
             }
         }
     }
 };

 int main()
 {
     vector<vector<int> > board;
     int i, j, n;
     Solution sol;

     while (scanf("%d", &n) ==  && n > ) {
         board.resize(n);
         for (i = ; i < n; ++i) {
             board[i].resize(n);
         }

         for (i = ; i < n; ++i) {
             for (j = ; j < n; ++j) {
                 scanf("%d", &board[i][j]);
             }
         }
         sol.rotateImage90(board);
         for (i = ; i < n; ++i) {
             printf("%d", board[i][]);
             for (j = ; j < n; ++j) {
                 printf(" %d", board[i][j]);
             }
             printf("\n");
         }
         printf("\n");

         for (i = ; i < n; ++i) {
             board[i].clear();
         }
         board.clear();
     }

     return ;
 }