feat: update solutions to lc problems: No.2108,2109 (#3148)

* No.2108.Find First Palindromic String in the Array
* No.2109.Adding Spaces to a String

Author: yanglbme

solution/2100-2199/2108.Find First Palindromic String in the Array/README.md

@@ -66,11 +66,11 @@ tags:
 
 ### 方法一：模拟
 
-遍历数组 `words`，对于每个字符串 `w`，判断其是否为回文字符串，如果是，则返回 `w`，否则继续遍历。
+我们遍历数组 `words`，对于每个字符串 `w`，判断其是否为回文字符串，如果是，则返回 `w`，否则继续遍历。
 
 判断一个字符串是否为回文字符串，可以使用双指针，分别指向字符串的首尾，向中间移动，判断对应的字符是否相等。如果遍历完整个字符串，都没有发现不相等的字符，则该字符串为回文字符串。
 
-时间复杂度 $O(L)$，空间复杂度 $O(1)$，其中 $L$ 为数组 `words` 中所有字符串的长度之和。
+时间复杂度 $O(L)$，其中 $L$ 为数组 `words` 中所有字符串的长度之和。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -148,21 +148,7 @@ func firstPalindrome(words []string) string {
 
 ```ts
 function firstPalindrome(words: string[]): string {
-    for (const word of words) {
-        let left = 0;
-        let right = word.length - 1;
-        while (left < right) {
-            if (word[left] !== word[right]) {
-                break;
-            }
-            left++;
-            right--;
-        }
-        if (left >= right) {
-            return word;
-        }
-    }
-    return '';
+    return words.find(w => w === w.split('').reverse().join('')) || '';
 }
 ```
 
@@ -171,19 +157,9 @@ function firstPalindrome(words: string[]): string {
 ```rust
 impl Solution {
     pub fn first_palindrome(words: Vec<String>) -> String {
-        for word in words.iter() {
-            let s = word.as_bytes();
-            let mut left = 0;
-            let mut right = s.len() - 1;
-            while left < right {
-                if s[left] != s[right] {
-                    break;
-                }
-                left += 1;
-                right -= 1;
-            }
-            if left >= right {
-                return word.clone();
+        for w in words {
+            if w == w.chars().rev().collect::<String>() {
+                return w;
             }
         }
         String::new()
@@ -195,18 +171,17 @@ impl Solution {
 
 ```c
 char* firstPalindrome(char** words, int wordsSize) {
-    for (int i = 0; i < wordsSize; i++) {
-        int left = 0;
-        int right = strlen(words[i]) - 1;
-        while (left < right) {
-            if (words[i][left] != words[i][right]) {
-                break;
+    for (int i = 0; i < wordsSize; ++i) {
+        char* w = words[i];
+        int len = strlen(w);
+        bool ok = true;
+        for (int j = 0, k = len - 1; j < k && ok; ++j, --k) {
+            if (w[j] != w[k]) {
+                ok = false;
             }
-            left++;
-            right--;
         }
-        if (left >= right) {
-            return words[i];
+        if (ok) {
+            return w;
         }
     }
     return "";

solution/2100-2199/2108.Find First Palindromic String in the Array/README_EN.md

@@ -65,7 +65,13 @@ Note that "racecar" is also palindromic, but it is not the first.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+We iterate through the array `words`, for each string `w`, we determine if it is a palindrome. If it is, then we return `w`; otherwise, we continue to iterate.
+
+To determine if a string is a palindrome, we can use two pointers, one pointing to the start and the other to the end of the string, moving towards the center, and checking if the corresponding characters are equal. If, after traversing the entire string, no unequal characters are found, then the string is a palindrome.
+
+The time complexity is $O(L)$, where $L$ is the sum of the lengths of all strings in the array `words`. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -143,21 +149,7 @@ func firstPalindrome(words []string) string {
 
 ```ts
 function firstPalindrome(words: string[]): string {
-    for (const word of words) {
-        let left = 0;
-        let right = word.length - 1;
-        while (left < right) {
-            if (word[left] !== word[right]) {
-                break;
-            }
-            left++;
-            right--;
-        }
-        if (left >= right) {
-            return word;
-        }
-    }
-    return '';
+    return words.find(w => w === w.split('').reverse().join('')) || '';
 }
 ```
 
@@ -166,19 +158,9 @@ function firstPalindrome(words: string[]): string {
 ```rust
 impl Solution {
     pub fn first_palindrome(words: Vec<String>) -> String {
-        for word in words.iter() {
-            let s = word.as_bytes();
-            let mut left = 0;
-            let mut right = s.len() - 1;
-            while left < right {
-                if s[left] != s[right] {
-                    break;
-                }
-                left += 1;
-                right -= 1;
-            }
-            if left >= right {
-                return word.clone();
+        for w in words {
+            if w == w.chars().rev().collect::<String>() {
+                return w;
             }
         }
         String::new()
@@ -190,18 +172,17 @@ impl Solution {
 
 ```c
 char* firstPalindrome(char** words, int wordsSize) {
-    for (int i = 0; i < wordsSize; i++) {
-        int left = 0;
-        int right = strlen(words[i]) - 1;
-        while (left < right) {
-            if (words[i][left] != words[i][right]) {
-                break;
+    for (int i = 0; i < wordsSize; ++i) {
+        char* w = words[i];
+        int len = strlen(w);
+        bool ok = true;
+        for (int j = 0, k = len - 1; j < k && ok; ++j, --k) {
+            if (w[j] != w[k]) {
+                ok = false;
             }
-            left++;
-            right--;
         }
-        if (left >= right) {
-            return words[i];
+        if (ok) {
+            return w;
         }
     }
     return "";

solution/2100-2199/2108.Find First Palindromic String in the Array/Solution.c

@@ -1,16 +1,15 @@
 char* firstPalindrome(char** words, int wordsSize) {
-    for (int i = 0; i < wordsSize; i++) {
-        int left = 0;
-        int right = strlen(words[i]) - 1;
-        while (left < right) {
-            if (words[i][left] != words[i][right]) {
-                break;
+    for (int i = 0; i < wordsSize; ++i) {
+        char* w = words[i];
+        int len = strlen(w);
+        bool ok = true;
+        for (int j = 0, k = len - 1; j < k && ok; ++j, --k) {
+            if (w[j] != w[k]) {
+                ok = false;
             }
-            left++;
-            right--;
         }
-        if (left >= right) {
-            return words[i];
+        if (ok) {
+            return w;
         }
     }
     return "";

solution/2100-2199/2108.Find First Palindromic String in the Array/Solution.rs

@@ -1,18 +1,8 @@
 impl Solution {
     pub fn first_palindrome(words: Vec<String>) -> String {
-        for word in words.iter() {
-            let s = word.as_bytes();
-            let mut left = 0;
-            let mut right = s.len() - 1;
-            while left < right {
-                if s[left] != s[right] {
-                    break;
-                }
-                left += 1;
-                right -= 1;
-            }
-            if left >= right {
-                return word.clone();
+        for w in words {
+            if w == w.chars().rev().collect::<String>() {
+                return w;
             }
         }
         String::new()

solution/2100-2199/2108.Find First Palindromic String in the Array/Solution.ts

@@ -1,17 +1,3 @@
 function firstPalindrome(words: string[]): string {
-    for (const word of words) {
-        let left = 0;
-        let right = word.length - 1;
-        while (left < right) {
-            if (word[left] !== word[right]) {
-                break;
-            }
-            left++;
-            right--;
-        }
-        if (left >= right) {
-            return word;
-        }
-    }
-    return '';
+    return words.find(w => w === w.split('').reverse().join('')) || '';
 }

solution/2100-2199/2109.Adding Spaces to a String/README.md

@@ -82,9 +82,9 @@ tags:
 
 ### 方法一：双指针
 
-我们可以用双指针 $i$ 和 $j$ 分别指向字符串 $s$ 和数组 $spaces$ 的头部，然后从头到尾遍历字符串 $s$，当 $i$ 等于 $spaces[j]$ 时，我们往结果字符串中添加一个空格，然后 $j$ 自增 1。接下来，我们将 $s[i]$ 添加到结果字符串中，然后 $i$ 自增 1。继续这个过程，直到遍历完字符串 $s$。
+我们可以用双指针 $i$ 和 $j$ 分别指向字符串 $s$ 和数组 $\text{spaces}$ 的头部，然后从头到尾遍历字符串 $s$，当 $i$ 等于 $\text{spaces}[j]$ 时，我们往结果字符串中添加一个空格，然后 $j$ 自增 $1$。接下来，我们将 $s[i]$ 添加到结果字符串中，然后 $i$ 自增 $1$。继续这个过程，直到遍历完字符串 $s$。
 
-时间复杂度 $O(n + m)$，其中 $n$ 和 $m$ 分别是字符串 $s$ 和数组 $spaces$ 的长度。忽略答案的空间消耗，空间复杂度 $O(1)$。
+时间复杂度 $O(n + m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别是字符串 $s$ 和数组 $spaces$ 的长度。
 
 <!-- tabs:start -->
 
@@ -160,46 +160,20 @@ func addSpaces(s string, spaces []int) string {
 
 ```ts
 function addSpaces(s: string, spaces: number[]): string {
-    let ans = '';
+    const ans: string[] = [];
     for (let i = 0, j = 0; i < s.length; i++) {
-        if (j < spaces.length && i === spaces[j]) {
-            ans += ' ';
-            ++j;
+        if (i === spaces[j]) {
+            ans.push(' ');
+            j++;
         }
-        ans += s[i];
+        ans.push(s[i]);
     }
-    return ans;
+    return ans.join('');
 }
 ```
 
 <!-- tabs:end -->
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def addSpaces(self, s: str, spaces: List[int]) -> str:
-        ans = []
-        i, j = len(s) - 1, len(spaces) - 1
-        while i >= 0:
-            ans.append(s[i])
-            if j >= 0 and i == spaces[j]:
-                ans.append(' ')
-                j -= 1
-            i -= 1
-        return ''.join(ans[::-1])
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2100-2199/2109.Adding Spaces to a String/README_EN.md

@@ -76,7 +76,11 @@ We are also able to place spaces before the first character of the string.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Two Pointers
+
+We can use two pointers $i$ and $j$ to point to the beginning of the string $s$ and the array $\text{spaces}$, respectively. Then, we iterate through the string $s$ from the beginning to the end. When $i$ equals $\text{spaces}[j]$, we add a space to the result string, and then increment $j$ by $1$. Next, we add $s[i]$ to the result string, and then increment $i$ by $1$. We continue this process until we have iterated through the entire string $s$.
+
+The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$, where $n$ and $m$ are the lengths of the string $s$ and the array $spaces$, respectively.
 
 <!-- tabs:start -->
 
@@ -152,46 +156,20 @@ func addSpaces(s string, spaces []int) string {
 
 ```ts
 function addSpaces(s: string, spaces: number[]): string {
-    let ans = '';
+    const ans: string[] = [];
     for (let i = 0, j = 0; i < s.length; i++) {
-        if (j < spaces.length && i === spaces[j]) {
-            ans += ' ';
-            ++j;
+        if (i === spaces[j]) {
+            ans.push(' ');
+            j++;
         }
-        ans += s[i];
+        ans.push(s[i]);
     }
-    return ans;
+    return ans.join('');
 }
 ```
 
 <!-- tabs:end -->
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def addSpaces(self, s: str, spaces: List[int]) -> str:
-        ans = []
-        i, j = len(s) - 1, len(spaces) - 1
-        while i >= 0:
-            ans.append(s[i])
-            if j >= 0 and i == spaces[j]:
-                ans.append(' ')
-                j -= 1
-            i -= 1
-        return ''.join(ans[::-1])
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2100-2199/2109.Adding Spaces to a String/Solution.ts

@@ -1,11 +1,11 @@
 function addSpaces(s: string, spaces: number[]): string {
-    let ans = '';
+    const ans: string[] = [];
     for (let i = 0, j = 0; i < s.length; i++) {
-        if (j < spaces.length && i === spaces[j]) {
-            ans += ' ';
-            ++j;
+        if (i === spaces[j]) {
+            ans.push(' ');
+            j++;
         }
-        ans += s[i];
+        ans.push(s[i]);
     }
-    return ans;
+    return ans.join('');
 }