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feat: add solutions to lc problems: No.1880,1881 (#4042)
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@yanglbme
yanglbme authored Feb 7, 2025
1 parent 8ce5ad2 commit 73e5c69
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99 changes: 54 additions & 45 deletions solution/1800-1899/1880.Check if Word Equals Summation of Two Words/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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 -->

Expand All @@ -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)
```
Expand All @@ -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;
}
}
```
Expand All @@ -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)
}
Expand All @@ -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)
}
}
```
Expand All @@ -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);
}
```
Expand Down
103 changes: 56 additions & 47 deletions solution/1800-1899/1880.Check if Word Equals Summation of Two Words/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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.
Expand All @@ -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.
Expand All @@ -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 -->

Expand All @@ -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)
```
Expand All @@ -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;
}
}
```
Expand All @@ -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)
}
Expand All @@ -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)
}
}
```
Expand All @@ -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);
}
```
Expand Down
15 changes: 8 additions & 7 deletions solution/1800-1899/1880.Check if Word Equals Summation of Two Words/Solution.c
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Original file line number Diff line number Diff line change
@@ -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);
}
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