【第三轮】7.24-7.31 CareerCup 5.8

5.8 A monochrome screen is stored as a single array of bytes, allowing eight consecutive pixels to be stored in one byte. The screen has width w, where w is divisible by 8 (that is, no byte will be split across rows). The height of the screen, of course, can be derived from the length of the array and the width. Implement a function drawHorizontalLine(byte[] screen, int width, intxl, intx2, inty) which draws a horizontal line from (x 1, y) to (x2, y).
回复解法可以按照以下格式来

【解题思路】
【时间复杂度】
【空间复杂度】
【gist link】
---------------OPTional，如果觉得test case比较好，欢迎写出来分享----------------------
【test case】

Notice：
1、记得在程序注释中表明自己算法的时间、空间复杂度
2、代码难懂之处加注释
3、每道题目有对应的帖子，除了贴解法，欢迎讨论，集思广益
4、任何未尽之处，欢迎回报名帖提问，我会进一步作出修改

【解题思路】最笨的方法，分情况讨论，题不难容但易出错
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/Noahsark/475ced1f7807f07c6f2e

【解题思路】
set the middle blocks as 0xFF
treat the first block and last block differently because only set parts of it as 1


【时间复杂度】
O(x2 - x1)

【空间复杂度】
O(1)

【gist link】
https://gist.github.com/happyWinner/aace24b9511a40db764f

【解题思路】
Find which row x1, x2 belongs to =>
if x1, x2 in same byte, set bits between x1 and x2 to 1;
if x1, x2 in different byte, set the byte of x1, x2, then set all bytes between x1 and x2 to 1

PS: the byte type has range from -128(10000000) ~ 127(01111111), in order to fit the requirement for the problem, we need use int[] rather than byte[]

【时间复杂度】
O(width)

【空间复杂度】
O(1)

【gist link】
https://gist.github.com/chrislukkk/c5606ab2c161c69f0f37

---------------OPTional，如果觉得test case比较好，欢迎写出来分享----------------------
【test case】
Wrote a function to print the screen

【解题思路】
draw a rough line first, then clear the leading pixels and fill in the tailing pixels
【时间复杂度】
O(W)
【空间复杂度】
O(1)
【gist link】
https://gist.github.com/cd8fccd3bc997b159ee0

【解题思路】
same method as in the book. find the bytes and use mask to change their values;
【时间复杂度】
O(N)
【空间复杂度】
O(1)
【gist link】
https://gist.github.com/jyhjuzi/9588ff61cd446a9761a3

【解题思路】书
【时间复杂度】o(n)
【空间复杂度】o(1)
【gist link】https://gist.github.com/hilda8519/3035c75c8500c22dccf5

【解题思路】
1. Find the position of x1,y; x2,y in the screen, then set the bits between them as 1
2. Set the bytes between x1,x2 as 0xFF, then set the bits after x1 as 1, the bits before x2 but not in a whole bytes as 1
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/bitcpf/5a7b7b9f08f0d47b1b8d

【解题思路】
find the index for x1 and x2, if they are the same, just & two mask, if not, or the mask1 with first index and or the mask2 with the second index, and set all 1s for between the two indices.
【时间复杂度】
O(W)
【空间复杂度】
O(1)
【gist link】
https://gist.github.com/iwilbert/da3752677be59a63da20