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Commit a0aa109

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feat: add solutions to lcci problem: No.05.08 (doocs#2672)
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‎lcci/05.08.Draw Line/README.md‎

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## 解法
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### 方法一:位运算
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我们先算出 $x_1$ 和 $x_2$ 在结果数组中的位置,记为 $i$ 和 $j$。然后将 $i$ 到 $j$ 之间的元素置为 $-1$。
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如果 $x_1 \bmod 32 \neq 0,ドル我们需要将 $i$ 位置的元素的前 $x_1 \bmod 32$ 位置为 0ドル$。
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如果 $x_2 \bmod 32 \neq 31,ドル我们需要将 $j$ 位置的元素的后 31ドル - x_2 \bmod 32$ 位置为 0ドル$。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def drawLine(self, length: int, w: int, x1: int, x2: int, y: int) -> List[int]:
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ans = [0] * length
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i = (y * w + x1) // 32
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j = (y * w + x2) // 32
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for k in range(i, j + 1):
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ans[k] = -1
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ans[i] = (ans[i] & 0xFFFFFFFF) >> (x1 % 32) if x1 % 32 else -1
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ans[j] &= -0x80000000 >> (x2 % 32)
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return ans
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```
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```java
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class Solution {
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public int[] drawLine(int length, int w, int x1, int x2, int y) {
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int[] ans = new int[length];
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int i = (y * w + x1) / 32;
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int j = (y * w + x2) / 32;
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for (int k = i; k <= j; ++k) {
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ans[k] = -1;
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}
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ans[i] = ans[i] >>> (x1 % 32);
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ans[j] &= 0x80000000 >> (x2 % 32);
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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vector<int> drawLine(int length, int w, int x1, int x2, int y) {
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vector<int> ans(length);
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int i = (y * w + x1) / 32;
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int j = (y * w + x2) / 32;
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for (int k = i; k <= j; ++k) {
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ans[k] = -1;
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}
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ans[i] = ans[i] & unsigned(-1) >> (x1 % 32);
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ans[j] = ans[j] & unsigned(-1) << (31 - x2 % 32);
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return ans;
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}
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};
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```
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<!-- tabs:end -->
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<!-- end -->

‎lcci/05.08.Draw Line/README_EN.md‎

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## Solutions
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### Solution 1: Bit Manipulation
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First, we calculate the positions of $x_1$ and $x_2$ in the result array, denoted as $i$ and $j$. Then, we set the elements between $i$ and $j$ to $-1$.
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If $x_1 \bmod 32 \neq 0,ドル we need to set the first $x_1 \bmod 32$ bits of the element at position $i$ to 0ドル$.
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If $x_2 \bmod 32 \neq 31,ドル we need to set the last 31ドル - x_2 \bmod 32$ bits of the element at position $j$ to 0ドル$.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def drawLine(self, length: int, w: int, x1: int, x2: int, y: int) -> List[int]:
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ans = [0] * length
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i = (y * w + x1) // 32
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j = (y * w + x2) // 32
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for k in range(i, j + 1):
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ans[k] = -1
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ans[i] = (ans[i] & 0xFFFFFFFF) >> (x1 % 32) if x1 % 32 else -1
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ans[j] &= -0x80000000 >> (x2 % 32)
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return ans
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```
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```java
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class Solution {
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public int[] drawLine(int length, int w, int x1, int x2, int y) {
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int[] ans = new int[length];
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int i = (y * w + x1) / 32;
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int j = (y * w + x2) / 32;
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for (int k = i; k <= j; ++k) {
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ans[k] = -1;
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}
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ans[i] = ans[i] >>> (x1 % 32);
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ans[j] &= 0x80000000 >> (x2 % 32);
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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vector<int> drawLine(int length, int w, int x1, int x2, int y) {
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vector<int> ans(length);
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int i = (y * w + x1) / 32;
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int j = (y * w + x2) / 32;
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for (int k = i; k <= j; ++k) {
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ans[k] = -1;
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}
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ans[i] = ans[i] & unsigned(-1) >> (x1 % 32);
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ans[j] = ans[j] & unsigned(-1) << (31 - x2 % 32);
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return ans;
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}
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};
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```
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<!-- tabs:end -->
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<!-- end -->

‎lcci/05.08.Draw Line/Solution.cpp‎

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class Solution {
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public:
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vector<int> drawLine(int length, int w, int x1, int x2, int y) {
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vector<int> ans(length);
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int i = (y * w + x1) / 32;
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int j = (y * w + x2) / 32;
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for (int k = i; k <= j; ++k) {
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ans[k] = -1;
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}
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ans[i] = ans[i] & unsigned(-1) >> (x1 % 32);
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ans[j] = ans[j] & unsigned(-1) << (31 - x2 % 32);
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return ans;
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}
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};

‎lcci/05.08.Draw Line/Solution.java‎

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class Solution {
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public int[] drawLine(int length, int w, int x1, int x2, int y) {
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int[] ans = new int[length];
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int i = (y * w + x1) / 32;
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int j = (y * w + x2) / 32;
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for (int k = i; k <= j; ++k) {
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ans[k] = -1;
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}
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ans[i] = ans[i] >>> (x1 % 32);
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ans[j] &= 0x80000000 >> (x2 % 32);
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return ans;
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}
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}

‎lcci/05.08.Draw Line/Solution.py‎

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class Solution:
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def drawLine(self, length: int, w: int, x1: int, x2: int, y: int) -> List[int]:
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ans = [0] * length
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i = (y * w + x1) // 32
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j = (y * w + x2) // 32
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for k in range(i, j + 1):
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ans[k] = -1
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ans[i] = (ans[i] & 0xFFFFFFFF) >> (x1 % 32) if x1 % 32 else -1
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ans[j] &= -0x80000000 >> (x2 % 32)
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return ans

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