《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 < 2) {
            return;
        }
        int w, h;
        int tmp;
        
        w = n / 2;
        h = n - w;
        for (i  = 0; i < w; ++i) {
            for (j = 0; j < h; ++j) {
                tmp = board[i][j];
                board[i][j] = board[n - 1 - j][i];
                board[n - 1 - j][i] = board[n - 1 - i][n - 1 - j];
                board[n - 1 - i][n - 1 - j] = board[j][n - 1 - i];
                board[j][n - 1 - i] = tmp;
            }
        }
    }
};

int main()
{
    vector<vector<int> > board;
    int i, j, n;
    Solution sol;
    
    while (scanf("%d", &n) == 1 && n > 0) {
        board.resize(n);
        for (i = 0; i < n; ++i) {
            board[i].resize(n);
        }
        
        for (i = 0; i < n; ++i) {
            for (j = 0; j < n; ++j) {
                scanf("%d", &board[i][j]);
            }
        }
        sol.rotateImage90(board);
        for (i = 0; i < n; ++i) {
            printf("%d", board[i][0]);
            for (j = 1; j < n; ++j) {
                printf(" %d", board[i][j]);
            }
            printf("
");
        }
        printf("
");
        
        for (i = 0; i < n; ++i) {
            board[i].clear();
        }
        board.clear();
    }
    
    return 0;
}