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feat: add solutions to lc problem: No.3758 #4860

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126 changes: 122 additions & 4 deletions solution/3700-3799/3758.Convert Number Words to Digits/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -103,32 +103,150 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3758.Co

<!-- solution:start -->

### 方法一
### 方法一：枚举

我们首先将数字单词与对应的数字建立映射关系，记录在数组 $d$ 中，其中 $d[i]$ 表示数字 $i$ 对应的单词。

然后我们从左到右遍历字符串 $s$，对于每个位置 $i$，我们依次枚举数字单词 $d[j]$，判断从位置 $i$ 开始的子串是否与 $d[j]$ 匹配。如果匹配成功，则将数字 $j$ 添加到结果中，并将位置 $i$ 向后移动 $|d[j]|$ 个位置。否则，将位置 $i$ 向后移动 1 个位置。

我们重复上述过程，直到遍历完整个字符串 $s$。最后将结果中的数字连接成字符串并返回。

时间复杂度 $O(n \times |d|)$，空间复杂度 $O(|d|)$，其中 $n$ 是字符串 $s$ 的长度，而 $|d|$ 是数字单词的数量。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def convertNumber(self, s: str) -> str:
d = [
"zero",
"one",
"two",
"three",
"four",
"five",
"six",
"seven",
"eight",
"nine",
]
i, n = 0, len(s)
ans = []
while i < n:
for j, t in enumerate(d):
m = len(t)
if i + m <= n and s[i : i + m] == t:
ans.append(str(j))
i += m - 1
break
i += 1
return "".join(ans)
```

#### Java

```java

class Solution {
public String convertNumber(String s) {
String[] d
= {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int n = s.length();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < d.length; ++j) {
String t = d[j];
int m = t.length();
if (i + m <= n && s.substring(i, i + m).equals(t)) {
ans.append(j);
i += m - 1;
break;
}
}
}
return ans.toString();
}
}
```

#### C++

```cpp

class Solution {
public:
string convertNumber(string s) {
vector<string> d = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int n = s.length();
string ans;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < d.size(); ++j) {
string t = d[j];
int m = t.length();
if (i + m <= n && s.substr(i, m) == t) {
ans += to_string(j);
i += m - 1;
break;
}
}
}
return ans;
}
};
```

#### Go

```go
func convertNumber(s string) string {
d := []string{"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}
n := len(s)
var ans strings.Builder
for i := 0; i < n; i++ {
for j, t := range d {
m := len(t)
if i+m <= n && s[i:i+m] == t {
ans.WriteString(strconv.Itoa(j))
i += m - 1
break
}
}
}
return ans.String()
}
```

#### TypeScript

```ts
function convertNumber(s: string): string {
const d: string[] = [
'zero',
'one',
'two',
'three',
'four',
'five',
'six',
'seven',
'eight',
'nine',
];
const n = s.length;
const ans: string[] = [];
for (let i = 0; i < n; ++i) {
for (let j = 0; j < d.length; ++j) {
const t = d[j];
const m = t.length;
if (i + m <= n && s.substring(i, i + m) === t) {
ans.push(j.toString());
i += m - 1;
break;
}
}
}
return ans.join('');
}
```

<!-- tabs:end -->
Expand Down
126 changes: 122 additions & 4 deletions solution/3700-3799/3758.Convert Number Words to Digits/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -103,32 +103,150 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3758.Co

<!-- solution:start -->

### Solution 1
### Solution 1: Enumeration

We first establish a mapping relationship between number words and their corresponding digits, recorded in array $d$, where $d[i]$ represents the word corresponding to digit $i$.

Then we traverse the string $s$ from left to right. For each position $i$, we enumerate the number words $d[j]$ in order and check whether the substring starting from position $i$ matches $d[j]$. If a match is found, we add digit $j$ to the result and move position $i$ forward by $|d[j]|$ positions. Otherwise, we move position $i$ forward by 1 position.

We repeat this process until we have traversed the entire string $s$. Finally, we concatenate the digits in the result into a string and return it.

The time complexity is $O(n \times |d|)$ and the space complexity is $O(|d|)$, where $n$ is the length of string $s$ and $|d|$ is the number of digit words.

<!-- tabs:start -->

#### Python3

```python

class Solution:
def convertNumber(self, s: str) -> str:
d = [
"zero",
"one",
"two",
"three",
"four",
"five",
"six",
"seven",
"eight",
"nine",
]
i, n = 0, len(s)
ans = []
while i < n:
for j, t in enumerate(d):
m = len(t)
if i + m <= n and s[i : i + m] == t:
ans.append(str(j))
i += m - 1
break
i += 1
return "".join(ans)
```

#### Java

```java

class Solution {
public String convertNumber(String s) {
String[] d
= {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int n = s.length();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < d.length; ++j) {
String t = d[j];
int m = t.length();
if (i + m <= n && s.substring(i, i + m).equals(t)) {
ans.append(j);
i += m - 1;
break;
}
}
}
return ans.toString();
}
}
```

#### C++

```cpp

class Solution {
public:
string convertNumber(string s) {
vector<string> d = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int n = s.length();
string ans;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < d.size(); ++j) {
string t = d[j];
int m = t.length();
if (i + m <= n && s.substr(i, m) == t) {
ans += to_string(j);
i += m - 1;
break;
}
}
}
return ans;
}
};
```

#### Go

```go
func convertNumber(s string) string {
d := []string{"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}
n := len(s)
var ans strings.Builder
for i := 0; i < n; i++ {
for j, t := range d {
m := len(t)
if i+m <= n && s[i:i+m] == t {
ans.WriteString(strconv.Itoa(j))
i += m - 1
break
}
}
}
return ans.String()
}
```

#### TypeScript

```ts
function convertNumber(s: string): string {
const d: string[] = [
'zero',
'one',
'two',
'three',
'four',
'five',
'six',
'seven',
'eight',
'nine',
];
const n = s.length;
const ans: string[] = [];
for (let i = 0; i < n; ++i) {
for (let j = 0; j < d.length; ++j) {
const t = d[j];
const m = t.length;
if (i + m <= n && s.substring(i, i + m) === t) {
ans.push(j.toString());
i += m - 1;
break;
}
}
}
return ans.join('');
}
```

<!-- tabs:end -->
Expand Down
20 changes: 20 additions & 0 deletions solution/3700-3799/3758.Convert Number Words to Digits/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
class Solution {
public:
string convertNumber(string s) {
vector<string> d = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int n = s.length();
string ans;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < d.size(); ++j) {
string t = d[j];
int m = t.length();
if (i + m <= n && s.substr(i, m) == t) {
ans += to_string(j);
i += m - 1;
break;
}
}
}
return ans;
}
};
16 changes: 16 additions & 0 deletions solution/3700-3799/3758.Convert Number Words to Digits/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
func convertNumber(s string) string {
d := []string{"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}
n := len(s)
var ans strings.Builder
for i := 0; i < n; i++ {
for j, t := range d {
m := len(t)
if i+m <= n && s[i:i+m] == t {
ans.WriteString(strconv.Itoa(j))
i += m - 1
break
}
}
}
return ans.String()
}
20 changes: 20 additions & 0 deletions solution/3700-3799/3758.Convert Number Words to Digits/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
class Solution {
public String convertNumber(String s) {
String[] d
= {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int n = s.length();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < d.length; ++j) {
String t = d[j];
int m = t.length();
if (i + m <= n && s.substring(i, i + m).equals(t)) {
ans.append(j);
i += m - 1;
break;
}
}
}
return ans.toString();
}
}
25 changes: 25 additions & 0 deletions solution/3700-3799/3758.Convert Number Words to Digits/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,25 @@
class Solution:
def convertNumber(self, s: str) -> str:
d = [
"zero",
"one",
"two",
"three",
"four",
"five",
"six",
"seven",
"eight",
"nine",
]
i, n = 0, len(s)
ans = []
while i < n:
for j, t in enumerate(d):
m = len(t)
if i + m <= n and s[i : i + m] == t:
ans.append(str(j))
i += m - 1
break
i += 1
return "".join(ans)
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