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May 27, 2024
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feat: add solutions to lc problem: No.3163 #2929

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105 changes: 101 additions & 4 deletions solution/3100-3199/3163.String Compression III/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -80,32 +80,129 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3163.St

<!-- solution:start -->

### 方法一
### 方法一：双指针

我们可以利用双指针，统计每个字符的连续出现次数。假如当前字符 $c$ 连续出现了 $k$ 次，然后我们将 $k$ 划分成若干个 $x$，每个 $x$ 最大为 $9$，然后将 $x$ 和 $c$ 拼接起来，将每个 $x$ 和 $c$ 拼接起来到结果中。

最后返回结果即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 `word` 的长度。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def compressedString(self, word: str) -> str:
g = groupby(word)
ans = []
for c, v in g:
k = len(list(v))
while k:
x = min(9, k)
ans.append(str(x) + c)
k -= x
return "".join(ans)
```

#### Java

```java

class Solution {
public String compressedString(String word) {
StringBuilder ans = new StringBuilder();
int n = word.length();
for (int i = 0; i < n;) {
int j = i + 1;
while (j < n && word.charAt(j) == word.charAt(i)) {
++j;
}
int k = j - i;
while (k > 0) {
int x = Math.min(9, k);
ans.append(x).append(word.charAt(i));
k -= x;
}
i = j;
}
return ans.toString();
}
}
```

#### C++

```cpp

class Solution {
public:
string compressedString(string word) {
string ans;
int n = word.length();
for (int i = 0; i < n;) {
int j = i + 1;
while (j < n && word[j] == word[i]) {
++j;
}
int k = j - i;
while (k > 0) {
int x = min(9, k);
ans.push_back('0' + x);
ans.push_back(word[i]);
k -= x;
}
i = j;
}
return ans;
}
};
```

#### Go

```go
func compressedString(word string) string {
ans := []byte{}
n := len(word)
for i := 0; i < n; {
j := i + 1
for j < n && word[j] == word[i] {
j++
}
k := j - i
for k > 0 {
x := min(9, k)
ans = append(ans, byte('0'+x))
ans = append(ans, word[i])
k -= x
}
i = j
}
return string(ans)
}
```

#### TypeScript

```ts
function compressedString(word: string): string {
const ans: string[] = [];
const n = word.length;
for (let i = 0; i < n; ) {
let j = i + 1;
while (j < n && word[j] === word[i]) {
++j;
}
let k = j - i;
while (k) {
const x = Math.min(k, 9);
ans.push(x + word[i]);
k -= x;
}
i = j;
}
return ans.join('');
}
```

<!-- tabs:end -->
Expand Down
105 changes: 101 additions & 4 deletions solution/3100-3199/3163.String Compression III/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -76,32 +76,129 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3163.St

<!-- solution:start -->

### Solution 1
### Solution 1: Two Pointers

We can use two pointers to count the consecutive occurrences of each character. Suppose the current character $c$ appears consecutively $k$ times, then we divide $k$ into several $x$, each $x$ is at most $9$, then we concatenate $x$ and $c$, and append each $x$ and $c$ to the result.

Finally, return the result.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the

<!-- tabs:start -->

#### Python3

```python

class Solution:
def compressedString(self, word: str) -> str:
g = groupby(word)
ans = []
for c, v in g:
k = len(list(v))
while k:
x = min(9, k)
ans.append(str(x) + c)
k -= x
return "".join(ans)
```

#### Java

```java

class Solution {
public String compressedString(String word) {
StringBuilder ans = new StringBuilder();
int n = word.length();
for (int i = 0; i < n;) {
int j = i + 1;
while (j < n && word.charAt(j) == word.charAt(i)) {
++j;
}
int k = j - i;
while (k > 0) {
int x = Math.min(9, k);
ans.append(x).append(word.charAt(i));
k -= x;
}
i = j;
}
return ans.toString();
}
}
```

#### C++

```cpp

class Solution {
public:
string compressedString(string word) {
string ans;
int n = word.length();
for (int i = 0; i < n;) {
int j = i + 1;
while (j < n && word[j] == word[i]) {
++j;
}
int k = j - i;
while (k > 0) {
int x = min(9, k);
ans.push_back('0' + x);
ans.push_back(word[i]);
k -= x;
}
i = j;
}
return ans;
}
};
```

#### Go

```go
func compressedString(word string) string {
ans := []byte{}
n := len(word)
for i := 0; i < n; {
j := i + 1
for j < n && word[j] == word[i] {
j++
}
k := j - i
for k > 0 {
x := min(9, k)
ans = append(ans, byte('0'+x))
ans = append(ans, word[i])
k -= x
}
i = j
}
return string(ans)
}
```

#### TypeScript

```ts
function compressedString(word: string): string {
const ans: string[] = [];
const n = word.length;
for (let i = 0; i < n; ) {
let j = i + 1;
while (j < n && word[j] === word[i]) {
++j;
}
let k = j - i;
while (k) {
const x = Math.min(k, 9);
ans.push(x + word[i]);
k -= x;
}
i = j;
}
return ans.join('');
}
```

<!-- tabs:end -->
Expand Down
22 changes: 22 additions & 0 deletions solution/3100-3199/3163.String Compression III/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -0,0 +1,22 @@
class Solution {
public:
string compressedString(string word) {
string ans;
int n = word.length();
for (int i = 0; i < n;) {
int j = i + 1;
while (j < n && word[j] == word[i]) {
++j;
}
int k = j - i;
while (k > 0) {
int x = min(9, k);
ans.push_back('0' + x);
ans.push_back(word[i]);
k -= x;
}
i = j;
}
return ans;
}
};
19 changes: 19 additions & 0 deletions solution/3100-3199/3163.String Compression III/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
func compressedString(word string) string {
ans := []byte{}
n := len(word)
for i := 0; i < n; {
j := i + 1
for j < n && word[j] == word[i] {
j++
}
k := j - i
for k > 0 {
x := min(9, k)
ans = append(ans, byte('0'+x))
ans = append(ans, word[i])
k -= x
}
i = j
}
return string(ans)
}
20 changes: 20 additions & 0 deletions solution/3100-3199/3163.String Compression III/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 compressedString(String word) {
StringBuilder ans = new StringBuilder();
int n = word.length();
for (int i = 0; i < n;) {
int j = i + 1;
while (j < n && word.charAt(j) == word.charAt(i)) {
++j;
}
int k = j - i;
while (k > 0) {
int x = Math.min(9, k);
ans.append(x).append(word.charAt(i));
k -= x;
}
i = j;
}
return ans.toString();
}
}
11 changes: 11 additions & 0 deletions solution/3100-3199/3163.String Compression III/Solution.py
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Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,11 @@
class Solution:
def compressedString(self, word: str) -> str:
g = groupby(word)
ans = []
for c, v in g:
k = len(list(v))
while k:
x = min(9, k)
ans.append(str(x) + c)
k -= x
return "".join(ans)
18 changes: 18 additions & 0 deletions solution/3100-3199/3163.String Compression III/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
function compressedString(word: string): string {
const ans: string[] = [];
const n = word.length;
for (let i = 0; i < n; ) {
let j = i + 1;
while (j < n && word[j] === word[i]) {
++j;
}
let k = j - i;
while (k) {
const x = Math.min(k, 9);
ans.push(x + word[i]);
k -= x;
}
i = j;
}
return ans.join('');
}
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