[LeetCode]48.Rotate Image

摘要： 【题目】 You are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? ...

【题目】

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

【题意】

给定一个n*n个2维矩阵来表示一个图。在原矩阵上旋转图形90°。

【分析】

思路1：

    

A[0][0] -> A[0][3]A[1][0] -> A[0][2]

A[0][1] -> A[1][3]A[2][0] -> A[0][1]

A[0][2] -> A[2][3]A[3][0] -> A[0][0]

A[0][3] -> A[3][3] 

由此可得：对于n * n 的2维矩阵

A[i][j] -> A[j][n-1-i]

思路2：

纯模拟，从外到内一圈一圈的转，但这个方法太慢。

A[i][j] -> A[j][n-1-i] -> A[n-1-i][n-1-j] -> A[n-1-j][i] -> A[i][j]

思路3：

【代码1】

/**********************************   日期：2014-01-21*   作者：SJF0115*   题号: Rotate Image*   来源：http://oj.leetcode.com/problems/rotate-image/*   结果：AC*   来源：LeetCode*   总结：**********************************/#include <iostream>#include <stdio.h>#include <malloc.h>#include <vector>using namespace std;class Solution {public:    void rotate(vector<vector<int> > &matrix) {        int i,j;        int n = matrix.size();        vector<vector<int> >tempMatrix = matrix;        for(i = 0;i < n;i++){            for(j = 0;j < n;j++){                tempMatrix[j][n-1-i] = matrix[i][j];            }//for        }//for        for(i = 0;i < n;i++){            for(j = 0;j < n;j++){                matrix[i][j] = tempMatrix[i][j];            }//for        }//for    }};int main() {    Solution solution;    vector<int> row1 = {1,2,3};    vector<int> row2 = {4,5,6};    vector<int> row3 = {7,8,9};    vector<vector<int>> matrix;    matrix.push_back(row1);    matrix.push_back(row2);    matrix.push_back(row3);    solution.rotate(matrix);    int n = matrix.size();    for(int i = 0;i < n;i++){        for(int j = 0;j < n;j++){            printf("%d ",matrix[i][j]);        }//for        printf("\n");    }//for    return 0;}

原地旋转，不能使用额外的空间存储矩阵。虽然本代码能AC，但是不符合题意。

【代码2】

class Solution {public:    void rotate(vector<vector<int> > &matrix) {        int i,j,temp;        int n=matrix.size();        for(i = 0;i < n/2;++i) {            for (j = i;j < n-1-i;++j) {                temp = matrix[j][n-i-1];                matrix[j][n-i-1] = matrix[i][j];                matrix[i][j] = matrix[n-j-1][i];                matrix[n-j-1][i] = matrix[n-i-1][n-j-1];                matrix[n-i-1][n-j-1] = temp;            }//for        }//for    }};

【代码3】

class Solution {public:    void rotate(vector<vector<int> > &matrix) {        int i,j,temp;        int n=matrix.size();        // 沿着副对角线反转        for (int i = 0; i < n; ++i) {            for (int j = 0; j < n - i; ++j) {                temp = matrix[i][j];                matrix[i][j] = matrix[n - 1 - j][n - 1 - i];                matrix[n - 1 - j][n - 1 - i] = temp;            }        }        // 沿着水平中线反转        for (int i = 0; i < n / 2; ++i){            for (int j = 0; j < n; ++j) {                temp = matrix[i][j];                matrix[i][j] = matrix[n - 1 - i][j];                matrix[n - 1 - i][j] = temp;            }        }    }};