feat: add solutions to lc problems: No.1880,1881 (#4042)

Author: yanglbme

solution/1800-1899/1880.Check if Word Equals Summation of Two Words/README.md

@@ -79,7 +79,13 @@ targetWord 的数值为 "aaaa" -> "0000" -> 0
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：字符串转数字
+
+我们定义一个函数 $\textit{f}(s)$，用来计算字符串 $s$ 的数值。对于字符串 $s$ 中的每个字符 $c$，我们将其转换为对应的数字 $x$，然后将 $x$ 依次连接起来，最后转换为整数。
+
+最后，我们只需要判断 $\textit{f}(\textit{firstWord}) + \textit{f}(\textit{secondWord})$ 是否等于 $\textit{f}(\textit{targetWord})$ 即可。
+
+时间复杂度 $O(L)$，其中 $L$ 为题目中所有字符串的长度之和。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -88,11 +94,12 @@ targetWord 的数值为 "aaaa" -&gt; "0000" -&gt; 0
 ```python
 class Solution:
     def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
-        def f(s):
-            res = 0
-            for c in s:
-                res = res * 10 + (ord(c) - ord('a'))
-            return res
+        def f(s: str) -> int:
+            ans, a = 0, ord("a")
+            for c in map(ord, s):
+                x = c - a
+                ans = ans * 10 + x
+            return ans
 
         return f(firstWord) + f(secondWord) == f(targetWord)
 ```
@@ -106,11 +113,11 @@ class Solution {
     }
 
     private int f(String s) {
-        int res = 0;
+        int ans = 0;
         for (char c : s.toCharArray()) {
-            res = res * 10 + (c - 'a');
+            ans = ans * 10 + (c - 'a');
         }
-        return res;
+        return ans;
     }
 }
 ```
@@ -121,27 +128,27 @@ class Solution {
 class Solution {
 public:
     bool isSumEqual(string firstWord, string secondWord, string targetWord) {
+        auto f = [](string& s) -> int {
+            int ans = 0;
+            for (char c : s) {
+                ans = ans * 10 + (c - 'a');
+            }
+            return ans;
+        };
         return f(firstWord) + f(secondWord) == f(targetWord);
     }
-
-    int f(string s) {
-        int res = 0;
-        for (char c : s) res = res * 10 + (c - 'a');
-        return res;
-    }
 };
 ```
 
 #### Go
 
 ```go
 func isSumEqual(firstWord string, secondWord string, targetWord string) bool {
-	f := func(s string) int {
-		res := 0
+	f := func(s string) (ans int) {
 		for _, c := range s {
-			res = res*10 + int(c-'a')
+			ans = ans*10 + int(c-'a')
 		}
-		return res
+		return
 	}
 	return f(firstWord)+f(secondWord) == f(targetWord)
 }
@@ -151,31 +158,32 @@ func isSumEqual(firstWord string, secondWord string, targetWord string) bool {
 
 ```ts
 function isSumEqual(firstWord: string, secondWord: string, targetWord: string): boolean {
-    const calc = (s: string) => {
-        let res = 0;
+    const f = (s: string): number => {
+        let ans = 0;
         for (const c of s) {
-            res = res * 10 + c.charCodeAt(0) - 'a'.charCodeAt(0);
+            ans = ans * 10 + c.charCodeAt(0) - 97;
         }
-        return res;
+        return ans;
     };
-    return calc(firstWord) + calc(secondWord) === calc(targetWord);
+    return f(firstWord) + f(secondWord) == f(targetWord);
 }
 ```
 
 #### Rust
 
 ```rust
 impl Solution {
-    fn calc(s: &String) -> i32 {
-        let mut res = 0;
-        for c in s.as_bytes() {
-            res = res * 10 + ((c - b'a') as i32);
-        }
-        res
-    }
-
     pub fn is_sum_equal(first_word: String, second_word: String, target_word: String) -> bool {
-        Self::calc(&first_word) + Self::calc(&second_word) == Self::calc(&target_word)
+        fn f(s: &str) -> i64 {
+            let mut ans = 0;
+            let a = 'a' as i64;
+            for c in s.chars() {
+                let x = c as i64 - a;
+                ans = ans * 10 + x;
+            }
+            ans
+        }
+        f(&first_word) + f(&second_word) == f(&target_word)
     }
 }
 ```
@@ -190,30 +198,31 @@ impl Solution {
  * @return {boolean}
  */
 var isSumEqual = function (firstWord, secondWord, targetWord) {
-    function f(s) {
-        let res = 0;
-        for (let c of s) {
-            res = res * 10 + (c.charCodeAt() - 'a'.charCodeAt());
+    const f = s => {
+        let ans = 0;
+        for (const c of s) {
+            ans = ans * 10 + c.charCodeAt(0) - 97;
         }
-        return res;
-    }
+        return ans;
+    };
     return f(firstWord) + f(secondWord) == f(targetWord);
 };
 ```
 
 #### C
 
 ```c
-int calc(char* s) {
-    int res = 0;
-    for (int i = 0; s[i]; i++) {
-        res = res * 10 + s[i] - 'a';
+int f(const char* s) {
+    int ans = 0;
+    while (*s) {
+        ans = ans * 10 + (*s - 'a');
+        s++;
     }
-    return res;
+    return ans;
 }
 
 bool isSumEqual(char* firstWord, char* secondWord, char* targetWord) {
-    return calc(firstWord) + calc(secondWord) == calc(targetWord);
+    return f(firstWord) + f(secondWord) == f(targetWord);
 }
 ```
 

solution/1800-1899/1880.Check if Word Equals Summation of Two Words/README_EN.md

@@ -48,7 +48,7 @@ We return true because 21 + 210 == 231.
 <pre>
 <strong>Input:</strong> firstWord = &quot;aaa&quot;, secondWord = &quot;a&quot;, targetWord = &quot;aab&quot;
 <strong>Output:</strong> false
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 The numerical value of firstWord is &quot;aaa&quot; -&gt; &quot;000&quot; -&gt; 0.
 The numerical value of secondWord is &quot;a&quot; -&gt; &quot;0&quot; -&gt; 0.
 The numerical value of targetWord is &quot;aab&quot; -&gt; &quot;001&quot; -&gt; 1.
@@ -60,7 +60,7 @@ We return false because 0 + 0 != 1.
 <pre>
 <strong>Input:</strong> firstWord = &quot;aaa&quot;, secondWord = &quot;a&quot;, targetWord = &quot;aaaa&quot;
 <strong>Output:</strong> true
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 The numerical value of firstWord is &quot;aaa&quot; -&gt; &quot;000&quot; -&gt; 0.
 The numerical value of secondWord is &quot;a&quot; -&gt; &quot;0&quot; -&gt; 0.
 The numerical value of targetWord is &quot;aaaa&quot; -&gt; &quot;0000&quot; -&gt; 0.
@@ -81,7 +81,13 @@ We return true because 0 + 0 == 0.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: String to Number
+
+We define a function $\textit{f}(s)$ to calculate the numerical value of the string $s$. For each character $c$ in the string $s$, we convert it to the corresponding number $x$, then concatenate $x$ sequentially, and finally convert it to an integer.
+
+Finally, we just need to check whether $\textit{f}(\textit{firstWord}) + \textit{f}(\textit{secondWord})$ equals $\textit{f}(\textit{targetWord})$.
+
+The time complexity is $O(L)$, where $L$ is the sum of the lengths of all strings in the problem. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -90,11 +96,12 @@ We return true because 0 + 0 == 0.
 ```python
 class Solution:
     def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
-        def f(s):
-            res = 0
-            for c in s:
-                res = res * 10 + (ord(c) - ord('a'))
-            return res
+        def f(s: str) -> int:
+            ans, a = 0, ord("a")
+            for c in map(ord, s):
+                x = c - a
+                ans = ans * 10 + x
+            return ans
 
         return f(firstWord) + f(secondWord) == f(targetWord)
 ```
@@ -108,11 +115,11 @@ class Solution {
     }
 
     private int f(String s) {
-        int res = 0;
+        int ans = 0;
         for (char c : s.toCharArray()) {
-            res = res * 10 + (c - 'a');
+            ans = ans * 10 + (c - 'a');
         }
-        return res;
+        return ans;
     }
 }
 ```
@@ -123,27 +130,27 @@ class Solution {
 class Solution {
 public:
     bool isSumEqual(string firstWord, string secondWord, string targetWord) {
+        auto f = [](string& s) -> int {
+            int ans = 0;
+            for (char c : s) {
+                ans = ans * 10 + (c - 'a');
+            }
+            return ans;
+        };
         return f(firstWord) + f(secondWord) == f(targetWord);
     }
-
-    int f(string s) {
-        int res = 0;
-        for (char c : s) res = res * 10 + (c - 'a');
-        return res;
-    }
 };
 ```
 
 #### Go
 
 ```go
 func isSumEqual(firstWord string, secondWord string, targetWord string) bool {
-	f := func(s string) int {
-		res := 0
+	f := func(s string) (ans int) {
 		for _, c := range s {
-			res = res*10 + int(c-'a')
+			ans = ans*10 + int(c-'a')
 		}
-		return res
+		return
 	}
 	return f(firstWord)+f(secondWord) == f(targetWord)
 }
@@ -153,31 +160,32 @@ func isSumEqual(firstWord string, secondWord string, targetWord string) bool {
 
 ```ts
 function isSumEqual(firstWord: string, secondWord: string, targetWord: string): boolean {
-    const calc = (s: string) => {
-        let res = 0;
+    const f = (s: string): number => {
+        let ans = 0;
         for (const c of s) {
-            res = res * 10 + c.charCodeAt(0) - 'a'.charCodeAt(0);
+            ans = ans * 10 + c.charCodeAt(0) - 97;
         }
-        return res;
+        return ans;
     };
-    return calc(firstWord) + calc(secondWord) === calc(targetWord);
+    return f(firstWord) + f(secondWord) == f(targetWord);
 }
 ```
 
 #### Rust
 
 ```rust
 impl Solution {
-    fn calc(s: &String) -> i32 {
-        let mut res = 0;
-        for c in s.as_bytes() {
-            res = res * 10 + ((c - b'a') as i32);
-        }
-        res
-    }
-
     pub fn is_sum_equal(first_word: String, second_word: String, target_word: String) -> bool {
-        Self::calc(&first_word) + Self::calc(&second_word) == Self::calc(&target_word)
+        fn f(s: &str) -> i64 {
+            let mut ans = 0;
+            let a = 'a' as i64;
+            for c in s.chars() {
+                let x = c as i64 - a;
+                ans = ans * 10 + x;
+            }
+            ans
+        }
+        f(&first_word) + f(&second_word) == f(&target_word)
     }
 }
 ```
@@ -192,30 +200,31 @@ impl Solution {
  * @return {boolean}
  */
 var isSumEqual = function (firstWord, secondWord, targetWord) {
-    function f(s) {
-        let res = 0;
-        for (let c of s) {
-            res = res * 10 + (c.charCodeAt() - 'a'.charCodeAt());
+    const f = s => {
+        let ans = 0;
+        for (const c of s) {
+            ans = ans * 10 + c.charCodeAt(0) - 97;
         }
-        return res;
-    }
+        return ans;
+    };
     return f(firstWord) + f(secondWord) == f(targetWord);
 };
 ```
 
 #### C
 
 ```c
-int calc(char* s) {
-    int res = 0;
-    for (int i = 0; s[i]; i++) {
-        res = res * 10 + s[i] - 'a';
+int f(const char* s) {
+    int ans = 0;
+    while (*s) {
+        ans = ans * 10 + (*s - 'a');
+        s++;
     }
-    return res;
+    return ans;
 }
 
 bool isSumEqual(char* firstWord, char* secondWord, char* targetWord) {
-    return calc(firstWord) + calc(secondWord) == calc(targetWord);
+    return f(firstWord) + f(secondWord) == f(targetWord);
 }
 ```
 

solution/1800-1899/1880.Check if Word Equals Summation of Two Words/Solution.c

@@ -1,11 +1,12 @@
-int calc(char* s) {
-    int res = 0;
-    for (int i = 0; s[i]; i++) {
-        res = res * 10 + s[i] - 'a';
+int f(const char* s) {
+    int ans = 0;
+    while (*s) {
+        ans = ans * 10 + (*s - 'a');
+        s++;
     }
-    return res;
+    return ans;
 }
 
 bool isSumEqual(char* firstWord, char* secondWord, char* targetWord) {
-    return calc(firstWord) + calc(secondWord) == calc(targetWord);
-}
+    return f(firstWord) + f(secondWord) == f(targetWord);
+}