feat: update solutions to lc problems: No.518,520 (doocs#2501)

Author: acbin

solution/0500-0599/0518.Coin Change II/README.md

@@ -129,7 +129,7 @@ class Solution {
 public:
     int change(int amount, vector<int>& coins) {
         int m = coins.size(), n = amount;
-        int f[m + 1][n + 1];
+        unsigned f[m + 1][n + 1];
         memset(f, 0, sizeof(f));
         f[0][0] = 1;
         for (int i = 1; i <= m; ++i) {
@@ -220,7 +220,7 @@ class Solution {
 public:
     int change(int amount, vector<int>& coins) {
         int n = amount;
-        int f[n + 1];
+        unsigned f[n + 1];
         memset(f, 0, sizeof(f));
         f[0] = 1;
         for (int x : coins) {

solution/0500-0599/0518.Coin Change II/README_EN.md

@@ -122,7 +122,7 @@ class Solution {
 public:
     int change(int amount, vector<int>& coins) {
         int m = coins.size(), n = amount;
-        int f[m + 1][n + 1];
+        unsigned f[m + 1][n + 1];
         memset(f, 0, sizeof(f));
         f[0][0] = 1;
         for (int i = 1; i <= m; ++i) {
@@ -213,7 +213,7 @@ class Solution {
 public:
     int change(int amount, vector<int>& coins) {
         int n = amount;
-        int f[n + 1];
+        unsigned f[n + 1];
         memset(f, 0, sizeof(f));
         f[0] = 1;
         for (int x : coins) {

solution/0500-0599/0518.Coin Change II/Solution.cpp

@@ -2,7 +2,7 @@ class Solution {
 public:
     int change(int amount, vector<int>& coins) {
         int m = coins.size(), n = amount;
-        int f[m + 1][n + 1];
+        unsigned f[m + 1][n + 1];
         memset(f, 0, sizeof(f));
         f[0][0] = 1;
         for (int i = 1; i <= m; ++i) {

solution/0500-0599/0518.Coin Change II/Solution2.cpp

@@ -2,7 +2,7 @@ class Solution {
 public:
     int change(int amount, vector<int>& coins) {
         int n = amount;
-        int f[n + 1];
+        unsigned f[n + 1];
         memset(f, 0, sizeof(f));
         f[0] = 1;
         for (int x : coins) {

solution/0500-0599/0520.Detect Capital/README.md

@@ -45,17 +45,22 @@
 
 ## 解法
 
-### 方法一
+### 方法一：统计大写字母的个数
+
+我们可以统计字符串中大写字母的个数，然后根据大写字母的个数判断是否符合题目要求。
+
+-   如果大写字母的个数为 0 或者等于字符串的长度，那么返回 `true`。
+-   如果大写字母的个数为 1 并且第一个字母是大写字母，那么返回 `true`。
+-   否则返回 `false`。
+
+时间复杂度 $O(n)$，其中 $n$ 为字符串 `word` 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def detectCapitalUse(self, word: str) -> bool:
-        cnt = 0
-        for c in word:
-            if c.isupper():
-                cnt += 1
+        cnt = sum(c.isupper() for c in word)
         return cnt == 0 or cnt == len(word) or (cnt == 1 and word[0].isupper())
 ```
 
@@ -78,10 +83,8 @@ class Solution {
 class Solution {
 public:
     bool detectCapitalUse(string word) {
-        int cnt = 0;
-        for (char c : word)
-            if (isupper(c)) ++cnt;
-        return cnt == 0 || cnt == word.size() || (cnt == 1 && isupper(word[0]));
+        int cnt = count_if(word.begin(), word.end(), [](char c) { return isupper(c); });
+        return cnt == 0 || cnt == word.length() || (cnt == 1 && isupper(word[0]));
     }
 };
 ```
@@ -98,6 +101,13 @@ func detectCapitalUse(word string) bool {
 }
 ```
 
+```ts
+function detectCapitalUse(word: string): boolean {
+    const cnt = word.split('').reduce((acc, c) => acc + (c === c.toUpperCase() ? 1 : 0), 0);
+    return cnt === 0 || cnt === word.length || (cnt === 1 && word[0] === word[0].toUpperCase());
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/0500-0599/0520.Detect Capital/README_EN.md

@@ -34,17 +34,22 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Count the Number of Uppercase Letters
+
+We can count the number of uppercase letters in the string, and then determine whether it meets the requirements of the problem based on the number of uppercase letters.
+
+-   If the number of uppercase letters is 0 or equal to the length of the string, then return `true`.
+-   If the number of uppercase letters is 1 and the first letter is an uppercase letter, then return `true`.
+-   Otherwise, return `false`.
+
+The time complexity is $O(n)$, where $n$ is the length of the string `word`. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def detectCapitalUse(self, word: str) -> bool:
-        cnt = 0
-        for c in word:
-            if c.isupper():
-                cnt += 1
+        cnt = sum(c.isupper() for c in word)
         return cnt == 0 or cnt == len(word) or (cnt == 1 and word[0].isupper())
 ```
 
@@ -67,10 +72,8 @@ class Solution {
 class Solution {
 public:
     bool detectCapitalUse(string word) {
-        int cnt = 0;
-        for (char c : word)
-            if (isupper(c)) ++cnt;
-        return cnt == 0 || cnt == word.size() || (cnt == 1 && isupper(word[0]));
+        int cnt = count_if(word.begin(), word.end(), [](char c) { return isupper(c); });
+        return cnt == 0 || cnt == word.length() || (cnt == 1 && isupper(word[0]));
     }
 };
 ```
@@ -87,6 +90,13 @@ func detectCapitalUse(word string) bool {
 }
 ```
 
+```ts
+function detectCapitalUse(word: string): boolean {
+    const cnt = word.split('').reduce((acc, c) => acc + (c === c.toUpperCase() ? 1 : 0), 0);
+    return cnt === 0 || cnt === word.length || (cnt === 1 && word[0] === word[0].toUpperCase());
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/0500-0599/0520.Detect Capital/Solution.cpp

@@ -1,9 +1,7 @@
 class Solution {
 public:
     bool detectCapitalUse(string word) {
-        int cnt = 0;
-        for (char c : word)
-            if (isupper(c)) ++cnt;
-        return cnt == 0 || cnt == word.size() || (cnt == 1 && isupper(word[0]));
+        int cnt = count_if(word.begin(), word.end(), [](char c) { return isupper(c); });
+        return cnt == 0 || cnt == word.length() || (cnt == 1 && isupper(word[0]));
     }
 };

solution/0500-0599/0520.Detect Capital/Solution.py

@@ -1,7 +1,4 @@
 class Solution:
     def detectCapitalUse(self, word: str) -> bool:
-        cnt = 0
-        for c in word:
-            if c.isupper():
-                cnt += 1
+        cnt = sum(c.isupper() for c in word)
         return cnt == 0 or cnt == len(word) or (cnt == 1 and word[0].isupper())

solution/0500-0599/0520.Detect Capital/Solution.ts

@@ -0,0 +1,4 @@
+function detectCapitalUse(word: string): boolean {
+    const cnt = word.split('').reduce((acc, c) => acc + (c === c.toUpperCase() ? 1 : 0), 0);
+    return cnt === 0 || cnt === word.length || (cnt === 1 && word[0] === word[0].toUpperCase());
+}