leetcode: rotate image

最新推荐文章于 2017-06-03 09:19:48 发布

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本文介绍了一种用于将n x n的二维矩阵顺时针旋转90度的算法，并提供了两种不同的实现方法。一种方法是通过直接交换元素来完成旋转；另一种方法则是先进行矩阵转置再进行水平翻转。

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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?

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#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
	void rotate(vector<vector<int> > &matrix) {
		/*
		 * time: 11ms, 6ms
		 * 直接找规则，对于每一个回字形，按顺时针方向交换左右上下四个数字，直接得到结果
		 */
		int n = matrix.size();
		int temp;
		for (int i = 0; i < n/2; i++)
		{
			for (int j = i; j < n-1-i; j++)
			{
				temp = matrix[i][j];
				matrix[i][j] = matrix[n-1-j][i];
				matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j];
				matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i];
				matrix[j][n - 1 - i] = temp;
			}
		}
	}

	void rotate2(vector<vector<int> > &matrix)
	{
		/* 先转换成转置矩阵，再左右翻转
		 * time ：7ms
		 */
		int n = matrix.size();
		int temp;
		for (int i = 0; i < n-1; i++)
		{
			for (int j = i + 1; j < n; j++)
			{
				temp = matrix[i][j];
				matrix[i][j] = matrix[j][i];
				matrix[j][i] = temp;
			}
		}
		for (int i = 0; i < n ; i++)
		{
			for (int j = 0; j < n/2; j++)
			{
				temp = matrix[i][j];
				matrix[i][j] = matrix[i][n-1-j];
				matrix[i][n - 1 - j] = temp;
			}
		}
	}
};
void main()
{
	vector<vector<int> > matrix;
	vector<int> row1 = { 1, 2, 3, 4 };
	vector<int> row2 = { 5, 6, 7, 8 };
	vector<int> row3 = { 9, 10, 11, 12 };
	vector<int> row4 = { 13, 14, 15, 16 };
	matrix.push_back(row1);
	matrix.push_back(row2);
	matrix.push_back(row3);
	matrix.push_back(row4);
	Solution s;
	s.rotate2(matrix);
	for (int i = 0; i < matrix.size(); i++)
	{
		for (int j = 0; j < matrix.size(); j++)
			cout << matrix[i][j] << "  ";
		cout << endl;
	}
}