leecode 解题总结：48. Rotate Image

#include <iostream>#include <stdio.h>#include <vector>#include <algorithm>using namespace std;/*问题: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?分析:这个是程序员面试经典的一道题目。题目本意是将n*n的矩阵顺时针旋转90度。参见下图，主要想法是：将111旋转到222的位置，222旋转到333的位置，333旋转444的位置，444旋转到111的位置.同理：将5旋转到6,6旋转到7,...1112456248724333问题的关键变成:如何确定上边到右边，右边到下边，下边到左边，左边到右边的坐标映射关系。上边到右边:特点,原先上边横坐标一致，变换到右边后，纵坐标一致。设置len(每次矩阵长度).offset:起点偏移值，从0开始,len + offset是最大纵坐标长度(0,0),(0,1),(0,2)， 变换成(0,3),(1,3),(2,3)所以上边(x,y),变换为右边(y,len + offset)右边变下边:(0,3),(1,3),(2,3) 变为(3,3),(3,2),(3,1),右边(x,len+offset)变成下边(len + offset, )下边变左边:(3,3),(3,2),(3,1) 变为(1,0),(2,0),(3,0)下边是左边变上边:(1,0),(2,0),(3,0) 变为(0,2),(0,1),(0,0)如果当前长度为n-1,下一次旋转的小矩阵长度为n-3(减去两边各1个)第一次旋转矩阵上边起始位置为:0~n-2,  二                        :1~n-3//控制循环次数int offset = -1;for(int len = n -1 ; len >= 0 ; len -= 2){    //决定左上角起点的横坐标offset++;for(int i = offset; i < offset + len ; i++){//进行变换处理}}正确变换顺序是: 临时=右边右边=上边上边=左边左边=下边下边=临时输入25 68 731 2 34 5 67 8 941 1 1 24 5 6 24 8 7 24 3 3 3输出:8 57 67 4 18 5 29 6 34 4 4 13 8 5 13 7 6 13 2 2 2关键:1 简单方法：图像旋转先从上到下每一行逆置，然后对称元素互换即可 * 1 2 3     7 8 9     7 4 1 * 4 5 6  => 4 5 6  => 8 5 2 * 7 8 9     1 2 3     9 6 32 旋转方法for(int len = n ; len > 0 ; len -= 2){//决定左上角起点的纵坐标，起始横坐标也一样。因为是矩阵offset++;border = offset + len - 1;count = 0;//这个边界有问题,len是矩阵长度:[0,3),[1,2)，最后一个元素无需反转for(int j = offset; j < border; j++){//临时 = 右边temp = matrix.at(j).at(border) ;//右边 = 上边, 右边: (0,3),(1,3),(2,3) (j,border)，上边:(0,0),(0,1),(0,2),即(offset,j) ,//右边:(1,2)，上边：(1,1)matrix.at(j).at(border) = matrix.at(offset).at(j);//上边 = 左边，上边:(0,0),(0,1),(0,2),即(offset,j) ，左边:(3,0),(2,0),(1,0),(border - count ,offset),//上边:(1,1)，左边:(2,1)，易错这里用count确保初始border-count一定是为边界，如果border-j就不是边界matrix.at(offset).at(j) = matrix.at(border - count).at(offset);// 左边 = 下边，左边:(3,0),(2,0),(1,0),(border -j,offset)，下边:(3,3),(3,2),(3,1)即(border , border -j)//左边:(2,1)，下边:(2,2)matrix.at(border - count).at(offset) =  matrix.at(border).at(border - count ) ;//下边 = 右边，下边:(3,3),(3,2),(3,1)，即(border , border - j)，matrix.at(border).at(border - count) = temp;count++;}}}*/void swap(int* num1 , int* num2){int temp = *num1;*num1 = *num2;*num2 = temp;}class Solution {public:    void rotate(vector<vector<int>>& matrix) {if(matrix.empty()){return ;}int n = matrix.size();//上下逆置reverse(matrix.begin() , matrix.end());//对称元素互换int size;for(int i = 0 ; i < n ; i++){size = matrix.at(i).size();for(int j = i + 1 ; j < size ; j++){swap(matrix[i][j] , matrix[j][i]);}}}    void rotate2(vector<vector<int>>& matrix) {if(matrix.empty()){return ;}int n = matrix.size();int offset = -1;int border;//边界int temp;int count;for(int len = n ; len > 0 ; len -= 2){//决定左上角起点的纵坐标，起始横坐标也一样。因为是矩阵offset++;border = offset + len - 1;count = 0;//这个边界有问题,len是矩阵长度:[0,3),[1,2)，最后一个元素无需反转for(int j = offset; j < border; j++){//临时 = 右边temp = matrix.at(j).at(border) ;//右边 = 上边, 右边: (0,3),(1,3),(2,3) (j,border)，上边:(0,0),(0,1),(0,2),即(offset,j) ,//右边:(1,2)，上边：(1,1)matrix.at(j).at(border) = matrix.at(offset).at(j);//上边 = 左边，上边:(0,0),(0,1),(0,2),即(offset,j) ，左边:(3,0),(2,0),(1,0),(border - count ,offset),//上边:(1,1)，左边:(2,1)，易错这里用count确保初始border-count一定是为边界，如果border-j就不是边界matrix.at(offset).at(j) = matrix.at(border - count).at(offset);// 左边 = 下边，左边:(3,0),(2,0),(1,0),(border -j,offset)，下边:(3,3),(3,2),(3,1)即(border , border -j)//左边:(2,1)，下边:(2,2)matrix.at(border - count).at(offset) =  matrix.at(border).at(border - count ) ;//下边 = 右边，下边:(3,3),(3,2),(3,1)，即(border , border - j)，matrix.at(border).at(border - count) = temp;count++;}}    }};void print(vector< vector<int> >& matrix){if(matrix.empty()){cout << "no result" << endl;return ;}int size = matrix.size();for(int i = 0 ; i < size ; i++){for(int j = 0 ; j < size ; j++){cout << matrix.at(i).at(j) << " ";}cout << endl;}}void process(){int num;int value;vector<vector<int> > matrix;Solution solution;while(cin >> num){matrix.clear();for(int i = 0 ; i < num ; i++){vector<int> nums;for(int j = 0 ; j < num ; j++){cin >> value;nums.push_back(value);}matrix.push_back(nums);}solution.rotate(matrix);print(matrix);}}int main(int argc , char* argv[]){process();getchar();return 0;}