• 博客园logo
  • 会员
  • 众包
  • 新闻
  • 博问
  • 闪存
  • 赞助商
  • HarmonyOS
  • Chat2DB
    • 搜索
      所有博客
    • 搜索
      当前博客
  • 写随笔 我的博客 短消息 简洁模式
    用户头像
    我的博客 我的园子 账号设置 会员中心 简洁模式 ... 退出登录
    注册 登录

nunca

但行好事 莫问前程
  • 博客园
  • 联系
  • 订阅
  • 管理

公告

View Post

LeetCode Easy: 48. Rotate Image

 一、题目

You are 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]
]

 题意:给定一个矩阵,顺时针旋转90度。
 二、思路

 

 三、代码

#coding:utf-8
import numpy as np
class Solution:
    def rotate(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: void Do not return anything, modify matrix in-place instead.
        """
        print("old:",'\n', matrix)
        n = np.shape(matrix)[0]
        # 沿着副对角线反转
        for i in range(n):
            for j in range(n-i):
                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 i in range(n//2):
            for j in range(n):
                temp = matrix[i][j]
                matrix[i][j] = matrix[n-1-i][j]
                matrix[n-1-i][j] = temp

        print('new:','\n', matrix)

if __name__ == '__main__':
    a = [[1,2,3],
         [4,5,6],
         [7,8,9]
        ]
    ss = Solution()
    ss.rotate(a)

  

既然无论如何时间都会过去,为什么不选择做些有意义的事情呢

posted on 2018-05-08 07:52  乐晓东随笔  阅读(178)  评论(0)    收藏  举报

刷新页面返回顶部
 
博客园  ©  2004-2025
浙公网安备 33010602011771号 浙ICP备2021040463号-3