[LeetCode]48.Rotate Image

【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;
}

【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;
            }
        }
    }
};