题目地址:https://leetcode.com/problems/rotate-image/description/

题目描述

Youare given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note: You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

题目大意

把矩阵原地顺时针旋转90度。

解题方法

做法挺简单,先上下翻转,再延左上到右下的对角线进行翻转(镜像操作)。

需要注意的是上下翻转的时候是rows-i-1,而不是rows-i。

代码:

class Solution(object):
    def rotate(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: void Do not return anything, modify matrix in-place instead.
        """
        if matrix:
            rows = len(matrix)
            cols = len(matrix[0])
            for i in xrange(rows / 2):
                for j in xrange(cols):
                    matrix[i][j], matrix[rows - i - 1][j] = matrix[rows - i - 1][j], matrix[i][j]
            for i in xrange(rows):
                for j in xrange(i):
                    matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]

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二刷的时候使用的另外一种旋转的方法:先左右镜像翻转,然后再沿着副对角线(从右上到左下的对角线↙)进行翻转。比上面的麻烦一点。

使用的C++代码如下:

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        if (!matrix.size()) return;
        const int N = matrix.size();
        for (int i = 0; i < N; i ++) {
            for (int j = 0; j < N / 2; j ++) {
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[i][N - j - 1];
                matrix[i][N - j - 1] = tmp;
            }
        }
        for (int i = 0; i < N; i ++) {
            for (int j = 0; j < N - i; j ++) {
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[N - j - 1][N - i - 1];
                matrix[N - j - 1][N - i - 1] = tmp;
            }
        }
    }
};

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