feat: add solutions to lcci problems: No.05.03,05.07 (doocs#2671)

yanglbme · web-flow · commit c993d20a751c · 2024-04-26T17:33:05.000+08:00
diff --git a/lcci/05.03.Reverse Bits/README.md b/lcci/05.03.Reverse Bits/README.md
@@ -91,6 +91,21 @@ func reverseBits(num int) (ans int) {
 }
 ```
 
+```ts
+function reverseBits(num: number): number {
+    let ans = 0;
+    let cnt = 0;
+    for (let i = 0, j = 0; i < 32; ++i) {
+        cnt += ((num >> i) & 1) ^ 1;
+        for (; cnt > 1; ++j) {
+            cnt -= ((num >> j) & 1) ^ 1;
+        }
+        ans = Math.max(ans, i - j + 1);
+    }
+    return ans;
+}
+```
+
 ```swift
 class Solution {
     func reverseBits(_ num: Int) -> Int {
diff --git a/lcci/05.03.Reverse Bits/README_EN.md b/lcci/05.03.Reverse Bits/README_EN.md
@@ -97,6 +97,21 @@ func reverseBits(num int) (ans int) {
 }
 ```
 
+```ts
+function reverseBits(num: number): number {
+    let ans = 0;
+    let cnt = 0;
+    for (let i = 0, j = 0; i < 32; ++i) {
+        cnt += ((num >> i) & 1) ^ 1;
+        for (; cnt > 1; ++j) {
+            cnt -= ((num >> j) & 1) ^ 1;
+        }
+        ans = Math.max(ans, i - j + 1);
+    }
+    return ans;
+}
+```
+
 ```swift
 class Solution {
     func reverseBits(_ num: Int) -> Int {
diff --git a/lcci/05.03.Reverse Bits/Solution.ts b/lcci/05.03.Reverse Bits/Solution.ts
@@ -0,0 +1,12 @@
+function reverseBits(num: number): number {
+    let ans = 0;
+    let cnt = 0;
+    for (let i = 0, j = 0; i < 32; ++i) {
+        cnt += ((num >> i) & 1) ^ 1;
+        for (; cnt > 1; ++j) {
+            cnt -= ((num >> j) & 1) ^ 1;
+        }
+        ans = Math.max(ans, i - j + 1);
+    }
+    return ans;
+}
diff --git a/lcci/05.06.Convert Integer/README.md b/lcci/05.06.Convert Integer/README.md
@@ -31,7 +31,7 @@
 
 ### 方法一：位运算
 
-我们将 A 和 B 进行异或运算，得到的结果中 $1$ 的个数即为需要改变的位数。
+我们将 A 和 B 进行异或运算，得到的结果的二进制表示中 $1$ 的个数即为需要改变的位数。
 
 时间复杂度 $O(\log n)$，其中 $n$ 为 A 和 B 的最大值。空间复杂度 $O(1)$。
 
diff --git a/lcci/05.07.Exchange/README.md b/lcci/05.07.Exchange/README.md
@@ -29,7 +29,11 @@
 
 ## 解法
 
-### 方法一
+### 方法一：位运算
+
+我们可以将 $\text{num}$ 与 $\text{0x55555555}$ 进行与运算，得到的结果是 $\text{num}$ 的偶数位，然后将其左移一位。再将 $\text{num}$ 与 $\text{0xaaaaaaaa}$ 进行与运算，得到的结果是 $\text{num}$ 的奇数位，然后将其右移一位。最后将两个结果进行或运算，即可得到答案。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -62,21 +66,17 @@ func exchangeBits(num int) int {
 }
 ```
 
+```ts
+function exchangeBits(num: number): number {
+    return ((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >>> 1);
+}
+```
+
 ```rust
 impl Solution {
-    pub fn exchange_bits(mut num: i32) -> i32 {
-        let mut res = 0;
-        let mut i = 0;
-        while num != 0 {
-            let a = num & 1;
-            num >>= 1;
-            let b = num & 1;
-            num >>= 1;
-            res |= a << (i + 1);
-            res |= b << i;
-            i += 2;
-        }
-        res
+    pub fn exchange_bits(num: i32) -> i32 {
+        let num = num as u32;
+        (((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >> 1)) as i32
     }
 }
 ```
diff --git a/lcci/05.07.Exchange/README_EN.md b/lcci/05.07.Exchange/README_EN.md
@@ -35,7 +35,11 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Bit Manipulation
+
+We can perform a bitwise AND operation between `num` and `0x55555555` to get the even bits of `num`, and then shift them one bit to the left. Then, we perform a bitwise AND operation between `num` and `0xaaaaaaaa` to get the odd bits of `num`, and then shift them one bit to the right. Finally, we perform a bitwise OR operation on the two results to get the answer.
+
+The time complexity is $O(1)$, and the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -68,21 +72,17 @@ func exchangeBits(num int) int {
 }
 ```
 
+```ts
+function exchangeBits(num: number): number {
+    return ((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >>> 1);
+}
+```
+
 ```rust
 impl Solution {
-    pub fn exchange_bits(mut num: i32) -> i32 {
-        let mut res = 0;
-        let mut i = 0;
-        while num != 0 {
-            let a = num & 1;
-            num >>= 1;
-            let b = num & 1;
-            num >>= 1;
-            res |= a << (i + 1);
-            res |= b << i;
-            i += 2;
-        }
-        res
+    pub fn exchange_bits(num: i32) -> i32 {
+        let num = num as u32;
+        (((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >> 1)) as i32
     }
 }
 ```
diff --git a/lcci/05.07.Exchange/Solution.rs b/lcci/05.07.Exchange/Solution.rs
@@ -1,16 +1,6 @@
 impl Solution {
-    pub fn exchange_bits(mut num: i32) -> i32 {
-        let mut res = 0;
-        let mut i = 0;
-        while num != 0 {
-            let a = num & 1;
-            num >>= 1;
-            let b = num & 1;
-            num >>= 1;
-            res |= a << (i + 1);
-            res |= b << i;
-            i += 2;
-        }
-        res
+    pub fn exchange_bits(num: i32) -> i32 {
+        let num = num as u32;
+        (((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >> 1)) as i32
     }
 }
diff --git a/lcci/05.07.Exchange/Solution.ts b/lcci/05.07.Exchange/Solution.ts
@@ -0,0 +1,3 @@
+function exchangeBits(num: number): number {
+    return ((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >>> 1);
+}
diff --git a/lcci/08.01.Three Steps Problem/README.md b/lcci/08.01.Three Steps Problem/README.md
@@ -200,6 +200,26 @@ class Solution:
         return sum(pow(a, n - 4)[0]) % mod
 ```
 
+```python
+import numpy as np
+
+
+class Solution:
+    def waysToStep(self, n: int) -> int:
+        if n < 4:
+            return 2 ** (n - 1)
+        mod = 10**9 + 7
+        factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
+        res = np.mat([(4, 2, 1)], np.dtype("O"))
+        n -= 4
+        while n:
+            if n & 1:
+                res = res * factor % mod
+            factor = factor * factor % mod
+            n >>= 1
+        return res.sum() % mod
+```
+
 ```java
 class Solution {
     private final int mod = (int) 1e9 + 7;
@@ -388,30 +408,4 @@ function pow(a, n) {
 
 <!-- tabs:end -->
 
-### 方法三
-
-<!-- tabs:start -->
-
-```python
-import numpy as np
-
-
-class Solution:
-    def waysToStep(self, n: int) -> int:
-        if n < 4:
-            return 2 ** (n - 1)
-        mod = 10**9 + 7
-        factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
-        res = np.mat([(4, 2, 1)], np.dtype("O"))
-        n -= 4
-        while n:
-            if n & 1:
-                res = res * factor % mod
-            factor = factor * factor % mod
-            n >>= 1
-        return res.sum() % mod
-```
-
-<!-- tabs:end -->
-
 <!-- end -->
diff --git a/lcci/08.01.Three Steps Problem/README_EN.md b/lcci/08.01.Three Steps Problem/README_EN.md
@@ -202,6 +202,26 @@ class Solution:
         return sum(pow(a, n - 4)[0]) % mod
 ```
 
+```python
+import numpy as np
+
+
+class Solution:
+    def waysToStep(self, n: int) -> int:
+        if n < 4:
+            return 2 ** (n - 1)
+        mod = 10**9 + 7
+        factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
+        res = np.mat([(4, 2, 1)], np.dtype("O"))
+        n -= 4
+        while n:
+            if n & 1:
+                res = res * factor % mod
+            factor = factor * factor % mod
+            n >>= 1
+        return res.sum() % mod
+```
+
 ```java
 class Solution {
     private final int mod = (int) 1e9 + 7;
@@ -390,30 +410,4 @@ function pow(a, n) {
 
 <!-- tabs:end -->
 
-### Solution 3
-
-<!-- tabs:start -->
-
-```python
-import numpy as np
-
-
-class Solution:
-    def waysToStep(self, n: int) -> int:
-        if n < 4:
-            return 2 ** (n - 1)
-        mod = 10**9 + 7
-        factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
-        res = np.mat([(4, 2, 1)], np.dtype("O"))
-        n -= 4
-        while n:
-            if n & 1:
-                res = res * factor % mod
-            factor = factor * factor % mod
-            n >>= 1
-        return res.sum() % mod
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,3 @@`
	`1`	`+function exchangeBits(num: number): number {`
	`2`	`+ return ((num & 0x55555555) << 1) \| ((num & 0xaaaaaaaa) >>> 1);`
	`3`	`+}`