LeetCode interview question 05.08. Draw a straight line

1. Title

  It is known that there is a monochrome screen composed of pixels. Each row has wpixels. The initial 0position of all pixels is (0,0).

  Now store the pixels of each 32row in a group according to "each pixel" int, and then store them lengthin a one-dimensional array with a length of .

(x1,y)We will draw a straight line from point to point   on the screen (x2,y)(that is, the pixel point is modified to 1), please return the drawn array.

Notice:

• The use case guarantees that the screen width wis evenly 32divisible by (i.e. one intwill not be spread over two lines)

Example 1:

Input: length = 1, w = 32, x1 = 30, x2 = 31, y = 0
Output: [3]
Explanation: Draw a straight line from the 30th to the 31st bit of line 0. The screen binary form is expressed as [ 00000000000000000000000000000011], so [3] is returned

Example 2:

Input: length = 3, w = 96, x1 = 0, x2 = 95, y = 0
Output: [-1, -1, -1]
Explanation: Since the int type of binary 11111111111111111111111111111111 represents -1, [-1 is returned ,-1,-1]

hint:

• 1 <= length <= 10^5
• 1 <= w <= 3 * 10^5
• 0 <= x1 <= x2 < w
• 0 <= y <= 10

  Click here to jump to the question .

2. Java problem solving

  After determining the left and right endpoints, assign values ​​based on the positions. The specific code is as follows:

class Solution {
    
    
    public int[] drawLine(int length, int w, int x1, int x2, int y) {
    
    
        int[] ans = new int[length];
        int col = w / 32, start = x1 / 32 + y * col, end = x2 / 32 + y * col;
        int left = (x1 % 32 == 0 ? 0 : 1 << (32 - x1 % 32)) - 1;
        int right = -(1 << (31 - x2 % 32));
        if (start == end) ans[start] = left + right + 1;
        else {
    
    
            ans[start] = left;
            for (int i = start + 1; i < end; i++) ans[i] = -1;
            ans[end] = right;
        }

        return ans;
    }
}

• Time: 0 ms, beats 100.00% of users using Java
• Memory: 40.81 MB, beats 95.08% of users using Java