Draw Line: 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 that draws a horizontal line from (x1, y) to (x2, y).
The method signature should look something like:
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drawline(byte[] screen, int width, int xl, int x2, int y)
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Hints: #366, #381, #384, #391
解法
byte[] screen是一个一维数组，里面存放的是一个byte，byte中的1bit代表一个像素点。里面存的是：
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| byte | byte | byte | byte |
[00100110,11100000,10100000,00001001, ...]
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width是8的倍数，单位是bit，height = screen * / width。
所谓画一条线就是把这条线上的像素点代表的bit设为1。注意这道题目中的画的是水平线，不是斜线，所以比较简单。
先把byte [] screen变成下面这种形式看看：
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|<- width ->|
00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000
(x1, y) (x2, y)
v v
00000000 00111111 11111111 11111000
00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000
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给定一个坐标让你求是第几个bit（从左往右数的，从0开始的）：
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int allBitIndex = (y - 1) * width + x
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求位于第几个byte：
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int byteIndex = allBitIndex / 8
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求位于这个byte里的第几个bit：
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int byteBitIndex = allBitIndex % 8
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代码：
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public void drawline(byte[] screen, int width, int xl, int x2, int y) {
Coord c1 = getCoord(width, x1, y);
Coord c2 = getCoord(width, x2, y);
byte allOne = (byte) (~0);
// 把x1, x2（不含）之间字节的都设1
for (int i = c1.byteIdx + 1; i < c2.byteIdx; i++) {
screen[c1.byteIdx] = allOne;
}
if (c1.byteIdx != c2.byteIdx) { // 两个像素在同一个byte里
// 把c1.bitIdx后面（含）都设1
screen[c1.byteIdx] |= allOne >> c1.bitIdx;
// 把c2.bitIdx前面（含）都设1
screen[c2.byteIdx] |= allOne << (8 - c2.bitIdx - 1);
} else {
screen[c1.byteIdx] |= (allOne >> c1.bitIdx) & (allOne << (8 - c2.bitIdx - 1));
}
}
private Coord getCoord(int width, int x, int y) {
int allBitIdx = (y - 1) * width + x;
int byteIdx = allBitIdx / 8; // 也可以是 allBitIdx >> 3
int bitIdx = allBitIdx % 8; // 也可以是 allBitIdx & 7 ( 0..0 111)
return new Coord(byteIdx, bitIdx);
}
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