旋转图像 · Rotate Image

[抄题]:

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]
]

 [暴力解法]:

时间分析:

空间分析:

[思维问题]:

[一句话思路]:

先xy翻转,再对折。就差一步了,观察力不够没看出来

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

[二刷]:

[三刷]:

[四刷]:

[五刷]:

  [五分钟肉眼debug的结果]:

[总结]:

i代表行的坐标,j代表列的坐标,j = i时代表xy 

[复杂度]:Time complexity: O(m*n) Space complexity: O(m*n)

[英文数据结构或算法,为什么不用别的数据结构或算法]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

289. Game of Life 题号小的很多题,就是一般的数组变换

 [代码风格] :

public class Solution {
    /*
     * @param matrix: a lists of integers
     * @return: 
     */
    public void rotate(int[][] matrix) {
        //corner case
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return ;
        }
        int length = matrix[0].length;
        //reverse
        for (int i = 0; i < matrix.length; i++) {
            for (int j = i; j < matrix[0].length; j++) {//xy
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
        //flip
        for (int i = 0; i < matrix.length; i++) {//all row
            for (int j = 0; j < matrix[0].length / 2; j++) {// half col
                int temp = matrix[i][j];
                matrix[i][j] = matrix[i][length - 1 - j];
                matrix[i][length - 1 - j] = temp;
            }
        }
    }
}
View Code

 

posted @ 2018-02-22 17:46  苗妙苗  阅读(230)  评论(0编辑  收藏  举报