【第三轮】6.16-6.22 CareerCup 1.6

【解题思路】preform a circular rotation on each layer
【时间复杂度】O(N^2)
【空间复杂度】O(1)
【gist link】https://gist.github.com/JoshuaTang/300b3e62fcc5449333a4

【解题思路】First we need to know the scale of N to determine use what type to restore this number in order not to overflow, since each pixel uses only 4 bytes, int is enough, we assume that the rotation is anticlockwise. We only need to rotate the upper-left quater "square", replace each quater square iterately, however, this is only the case when N is even, when N is odd, the square has overlap and some pixle might be rotate more than one time. The responding position of pixel in each quater "square" is (i, j), (j, n-i), (n-i, n-j) and (n-j, i), this could be obtained recursively.
【时间复杂度】O(N2)
【空间复杂度】O(1)
【gist link】
https://gist.github.com/6af17484b1fc52138a23.git
---------------OPTional，如果觉得test case比较好，欢迎写出来分享----------------------
【test case】

【解题思路】
rotate one layer by another, from outermost to innermost.

【时间复杂度】
O(n^2)


【空间复杂度】
O(1)


【gist link】
https://gist.github.com/happyWinner/cccb48c218c48b996fde