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 drawHorizontall_ine(byte[] screen, int width, int xl, int x2, int y) which draws a horizontal line from (xl, y)to(x2, y).
这道题给了我们一个字节数组,用来表示一个单色的屏幕,并给定我们两点坐标,让我们画一条线段。这让我想起了小学的时候,机房的那个电脑只能用图龟在屏幕上画线(呀,暴露年龄了-.-|||),当然那时候我不可能知道原理的。言归正传,这道题给我们的点的y坐标都相同,就是让我们画一条直线,大大降低了难度。当然我们可以按位来操作,但是这样的解题就不是出题者要考察的本意了,我们需要直接对byte处理。思路是首先算出起点和终点之间有多少字节是可以完全填充的,先把这些字节填充好,然后再分别处理开头和结尾的字节,参见代码如下:
class Solution {
public:
void drawLine(vector<unsigned char> &screen, int width, int x1, int x2, int y) {
int start_offset = x1 % , first_full_byte = x1 / ;
int end_offset = x2 % , last_full_byte = x2 / ;
if (start_offset != ) ++first_full_byte;
if (end_offset != ) --last_full_byte;
for (int i = first_full_byte; i <= last_full_byte; ++i) {
screen[(width / ) * y + i] = (unsigned char) 0xFF;
}
unsigned char start_mask = (unsigned char) 0xFF >> start_offset;
unsigned char end_mask = (unsigned char) 0xFF >> ( - end_offset);
if (start_offset != ) {
int byte_idx = (width / ) * y + first_full_byte - ;
screen[byte_idx] |= start_mask;
}
if (end_offset != ) {
int byte_idx = (width / ) * y + last_full_byte + ;
screen[byte_idx] |= end_mask;
}
}
};