feat: add solutions to lc problems: No.190~195 (doocs#2633)

Author: yanglbme

solution/0100-0199/0190.Reverse Bits/README.md

@@ -49,30 +49,36 @@
 
 ## 解法
 
-### 方法一
+### 方法一：位运算
+
+我们可以从低位到高位，逐个取出 `n` 的每一位，然后将其放到 `ans` 的对应位置上。
+
+比如，对于第 $i$ 位，我们可以通过 `(n & 1) << (31 - i)` 来取出 `n` 的第 $i$ 位，并将其放到 `ans` 的第 $31 - i$ 位上，然后将 `n` 右移一位。
+
+时间复杂度 $(\log n)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def reverseBits(self, n: int) -> int:
-        res = 0
+        ans = 0
         for i in range(32):
-            res |= (n & 1) << (31 - i)
+            ans |= (n & 1) << (31 - i)
             n >>= 1
-        return res
+        return ans
 ```
 
 ```java
 public class Solution {
     // you need treat n as an unsigned value
     public int reverseBits(int n) {
-        int res = 0;
+        int ans = 0;
         for (int i = 0; i < 32 && n != 0; ++i) {
-            res |= ((n & 1) << (31 - i));
+            ans |= (n & 1) << (31 - i);
             n >>>= 1;
         }
-        return res;
+        return ans;
     }
 }
 ```
@@ -81,36 +87,35 @@ public class Solution {
 class Solution {
 public:
     uint32_t reverseBits(uint32_t n) {
-        uint32_t res = 0;
-        for (int i = 0; i < 32; ++i) {
-            res |= ((n & 1) << (31 - i));
+        uint32_t ans = 0;
+        for (int i = 0; i < 32 && n; ++i) {
+            ans |= (n & 1) << (31 - i);
             n >>= 1;
         }
-        return res;
+        return ans;
     }
 };
 ```
 
 ```go
-func reverseBits(num uint32) uint32 {
-	var ans uint32 = 0
+func reverseBits(n uint32) (ans uint32) {
 	for i := 0; i < 32; i++ {
-		ans |= (num & 1) << (31 - i)
-		num >>= 1
+		ans |= (n & 1) << (31 - i)
+		n >>= 1
 	}
-	return ans
+	return
 }
 ```
 
 ```rust
 impl Solution {
-    pub fn reverse_bits(mut x: u32) -> u32 {
-        let mut res = 0;
-        for _ in 0..32 {
-            res = (res << 1) | (x & 1);
-            x >>= 1;
+    pub fn reverse_bits(mut n: u32) -> u32 {
+        let mut ans = 0;
+        for i in 0..32 {
+            ans |= (n & 1) << (31 - i);
+            n >>= 1;
         }
-        res
+        ans
     }
 }
 ```
@@ -121,13 +126,12 @@ impl Solution {
  * @return {number} - a positive integer
  */
 var reverseBits = function (n) {
-    let res = 0;
-    for (let i = 0; i < 32 && n > 0; ++i) {
-        res |= (n & 1) << (31 - i);
-        n >>>= 1;
+    let ans = 0;
+    for (let i = 0; i < 32 && n; ++i) {
+        ans |= (n & 1) << (31 - i);
+        n >>= 1;
     }
-    // 无符号右移
-    return res >>> 0;
+    return ans >>> 0;
 };
 ```
 

solution/0100-0199/0190.Reverse Bits/README_EN.md

@@ -44,30 +44,36 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Bit Manipulation
+
+We can extract each bit of `n` from the least significant bit to the most significant bit, and then place it in the corresponding position of `ans`.
+
+For example, for the $i$-th bit, we can use `(n & 1) << (31 - i)` to extract the $i$-th bit of `n` and place it on the $31 - i$-th bit of `ans`, then right shift `n` by one bit.
+
+The time complexity is $O(\log n)$, and the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def reverseBits(self, n: int) -> int:
-        res = 0
+        ans = 0
         for i in range(32):
-            res |= (n & 1) << (31 - i)
+            ans |= (n & 1) << (31 - i)
             n >>= 1
-        return res
+        return ans
 ```
 
 ```java
 public class Solution {
     // you need treat n as an unsigned value
     public int reverseBits(int n) {
-        int res = 0;
+        int ans = 0;
         for (int i = 0; i < 32 && n != 0; ++i) {
-            res |= ((n & 1) << (31 - i));
+            ans |= (n & 1) << (31 - i);
             n >>>= 1;
         }
-        return res;
+        return ans;
     }
 }
 ```
@@ -76,36 +82,35 @@ public class Solution {
 class Solution {
 public:
     uint32_t reverseBits(uint32_t n) {
-        uint32_t res = 0;
-        for (int i = 0; i < 32; ++i) {
-            res |= ((n & 1) << (31 - i));
+        uint32_t ans = 0;
+        for (int i = 0; i < 32 && n; ++i) {
+            ans |= (n & 1) << (31 - i);
             n >>= 1;
         }
-        return res;
+        return ans;
     }
 };
 ```
 
 ```go
-func reverseBits(num uint32) uint32 {
-	var ans uint32 = 0
+func reverseBits(n uint32) (ans uint32) {
 	for i := 0; i < 32; i++ {
-		ans |= (num & 1) << (31 - i)
-		num >>= 1
+		ans |= (n & 1) << (31 - i)
+		n >>= 1
 	}
-	return ans
+	return
 }
 ```
 
 ```rust
 impl Solution {
-    pub fn reverse_bits(mut x: u32) -> u32 {
-        let mut res = 0;
-        for _ in 0..32 {
-            res = (res << 1) | (x & 1);
-            x >>= 1;
+    pub fn reverse_bits(mut n: u32) -> u32 {
+        let mut ans = 0;
+        for i in 0..32 {
+            ans |= (n & 1) << (31 - i);
+            n >>= 1;
         }
-        res
+        ans
     }
 }
 ```
@@ -116,13 +121,12 @@ impl Solution {
  * @return {number} - a positive integer
  */
 var reverseBits = function (n) {
-    let res = 0;
-    for (let i = 0; i < 32 && n > 0; ++i) {
-        res |= (n & 1) << (31 - i);
-        n >>>= 1;
+    let ans = 0;
+    for (let i = 0; i < 32 && n; ++i) {
+        ans |= (n & 1) << (31 - i);
+        n >>= 1;
     }
-    // 无符号右移
-    return res >>> 0;
+    return ans >>> 0;
 };
 ```
 

solution/0100-0199/0190.Reverse Bits/Solution.cpp

@@ -1,11 +1,11 @@
 class Solution {
 public:
     uint32_t reverseBits(uint32_t n) {
-        uint32_t res = 0;
-        for (int i = 0; i < 32; ++i) {
-            res |= ((n & 1) << (31 - i));
+        uint32_t ans = 0;
+        for (int i = 0; i < 32 && n; ++i) {
+            ans |= (n & 1) << (31 - i);
             n >>= 1;
         }
-        return res;
+        return ans;
     }
 };

solution/0100-0199/0190.Reverse Bits/Solution.go

@@ -1,8 +1,7 @@
-func reverseBits(num uint32) uint32 {
-	var ans uint32 = 0
+func reverseBits(n uint32) (ans uint32) {
 	for i := 0; i < 32; i++ {
-		ans |= (num & 1) << (31 - i)
-		num >>= 1
+		ans |= (n & 1) << (31 - i)
+		n >>= 1
 	}
-	return ans
+	return
 }

solution/0100-0199/0190.Reverse Bits/Solution.java

@@ -1,11 +1,11 @@
 public class Solution {
     // you need treat n as an unsigned value
     public int reverseBits(int n) {
-        int res = 0;
+        int ans = 0;
         for (int i = 0; i < 32 && n != 0; ++i) {
-            res |= ((n & 1) << (31 - i));
+            ans |= (n & 1) << (31 - i);
             n >>>= 1;
         }
-        return res;
+        return ans;
     }
 }

solution/0100-0199/0190.Reverse Bits/Solution.js

@@ -3,11 +3,10 @@
  * @return {number} - a positive integer
  */
 var reverseBits = function (n) {
-    let res = 0;
-    for (let i = 0; i < 32 && n > 0; ++i) {
-        res |= (n & 1) << (31 - i);
-        n >>>= 1;
+    let ans = 0;
+    for (let i = 0; i < 32 && n; ++i) {
+        ans |= (n & 1) << (31 - i);
+        n >>= 1;
     }
-    // 无符号右移
-    return res >>> 0;
+    return ans >>> 0;
 };

solution/0100-0199/0190.Reverse Bits/Solution.py

@@ -1,7 +1,7 @@
 class Solution:
     def reverseBits(self, n: int) -> int:
-        res = 0
+        ans = 0
         for i in range(32):
-            res |= (n & 1) << (31 - i)
+            ans |= (n & 1) << (31 - i)
             n >>= 1
-        return res
+        return ans

solution/0100-0199/0190.Reverse Bits/Solution.rs

@@ -1,10 +1,10 @@
 impl Solution {
-    pub fn reverse_bits(mut x: u32) -> u32 {
-        let mut res = 0;
-        for _ in 0..32 {
-            res = (res << 1) | (x & 1);
-            x >>= 1;
+    pub fn reverse_bits(mut n: u32) -> u32 {
+        let mut ans = 0;
+        for i in 0..32 {
+            ans |= (n & 1) << (31 - i);
+            n >>= 1;
         }
-        res
+        ans
     }
 }

solution/0100-0199/0192.Word Frequency/README.md

@@ -45,4 +45,15 @@ day 1
 
 ## 解法
 
+### 方法一：awk
+
+<!-- tabs:start -->
+
+```bash
+# Read from the file words.txt and output the word frequency list to stdout.
+cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -nr | awk '{print $2, $1}'
+```
+
+<!-- tabs:end -->
+
 <!-- end -->

solution/0100-0199/0192.Word Frequency/README_EN.md

@@ -43,4 +43,15 @@ day 1
 
 ## Solutions
 
+### Solution 1: awk
+
+<!-- tabs:start -->
+
+```bash
+# Read from the file words.txt and output the word frequency list to stdout.
+cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -nr | awk '{print $2, $1}'
+```
+
+<!-- tabs:end -->
+
 <!-- end -->

solution/0100-0199/0193.Valid Phone Numbers/README.md

@@ -35,4 +35,15 @@
 
 ## 解法
 
+### 方法一：awk
+
+<!-- tabs:start -->
+
+```bash
+# Read from the file file.txt and output all valid phone numbers to stdout.
+awk '/^([0-9]{3}-|\([0-9]{3}\) )[0-9]{3}-[0-9]{4}$/' file.txt
+```
+
+<!-- tabs:end -->
+
 <!-- end -->

solution/0100-0199/0193.Valid Phone Numbers/README_EN.md

@@ -31,4 +31,15 @@
 
 ## Solutions
 
+### Solution 1: awk
+
+<!-- tabs:start -->
+
+```bash
+# Read from the file file.txt and output all valid phone numbers to stdout.
+awk '/^([0-9]{3}-|\([0-9]{3}\) )[0-9]{3}-[0-9]{4}$/' file.txt
+```
+
+<!-- tabs:end -->
+
 <!-- end -->