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feat: add solutions to lc problem: No.1957 #3547

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Sep 21, 2024
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feat: add solutions to lc problem: No.1957 #3547

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62 changes: 41 additions & 21 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -70,7 +70,13 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：模拟

我们可以遍历字符串 $s$，并使用一个数组 $\textit{ans}$ 记录当前的答案。对于每一个字符 $c$，如果 $\textit{ans}$ 的长度小于 $2$ 或者 $\textit{ans}$ 的最后两个字符不等于 $c$，我们就将 $c$ 添加到 $\textit{ans}$ 中。

最后，我们将 $\textit{ans}$ 中的字符连接起来，就得到了答案。

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

<!-- tabs:start -->

Expand All @@ -81,10 +87,9 @@ class Solution:
def makeFancyString(self, s: str) -> str:
ans = []
for c in s:
if len(ans) > 1 and ans[-1] == ans[-2] == c:
continue
ans.append(c)
return ''.join(ans)
if len(ans) < 2 or ans[-1] != c or ans[-2] != c:
ans.append(c)
return "".join(ans)
```

#### Java
Expand All @@ -95,10 +100,9 @@ class Solution {
StringBuilder ans = new StringBuilder();
for (char c : s.toCharArray()) {
int n = ans.length();
if (n > 1 && ans.charAt(n - 1) == c && ans.charAt(n - 2) == c) {
continue;
if (n < 2 || c != ans.charAt(n - 1) || c != ans.charAt(n - 2)) {
ans.append(c);
}
ans.append(c);
}
return ans.toString();
}
Expand All @@ -114,8 +118,9 @@ public:
string ans = "";
for (char& c : s) {
int n = ans.size();
if (n > 1 && ans[n - 1] == c && ans[n - 2] == c) continue;
ans.push_back(c);
if (n < 2 || ans[n - 1] != c || ans[n - 2] != c) {
ans += c;
}
}
return ans;
}
Expand All @@ -128,16 +133,29 @@ public:
func makeFancyString(s string) string {
ans := []rune{}
for _, c := range s {
n := len(ans)
if n > 1 && ans[n-1] == c && ans[n-2] == c {
continue
if n := len(ans); n < 2 || c != ans[n-1] || c != ans[n-2] {
ans = append(ans, c)
}
ans = append(ans, c)
}
return string(ans)
}
```

#### TypeScript

```ts
function makeFancyString(s: string): string {
const ans: string[] = [];
for (const c of s) {
const n = ans.length;
if (n < 2 || c !== ans[n - 1] || c !== ans[n - 2]) {
ans.push(c);
}
}
return ans.join('');
}
```

#### PHP

```php
Expand All @@ -147,15 +165,17 @@ class Solution {
* @return String
*/
function makeFancyString($s) {
$rs = '';
for ($i = 0; $i < strlen($s); $i++) {
if ($s[$i] == $s[$i + 1] && $s[$i] == $s[$i + 2]) {
continue;
} else {
$rs .= $s[$i];
$ans = [];
$length = strlen($s);

for ($i = 0; $i < $length; $i++) {
$n = count($ans);
if ($n < 2 || $s[$i] !== $ans[$n - 1] || $s[$i] !== $ans[$n - 2]) {
$ans[] = $s[$i];
}
}
return $rs;

return implode('', $ans);
}
}
```
Expand Down
62 changes: 41 additions & 21 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -68,7 +68,13 @@ No three consecutive characters are equal, so return &quot;aabaa&quot;.

<!-- solution:start -->

### Solution 1
### Solution 1: Simulation

We can traverse the string $s$ and use an array $\textit{ans}$ to record the current answer. For each character $c$, if the length of $\textit{ans}$ is less than $2$ or the last two characters of $\textit{ans}$ are not equal to $c$, we add $c$ to $\textit{ans}$.

Finally, we concatenate the characters in $\textit{ans}$ to get the answer.

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

<!-- tabs:start -->

Expand All @@ -79,10 +85,9 @@ class Solution:
def makeFancyString(self, s: str) -> str:
ans = []
for c in s:
if len(ans) > 1 and ans[-1] == ans[-2] == c:
continue
ans.append(c)
return ''.join(ans)
if len(ans) < 2 or ans[-1] != c or ans[-2] != c:
ans.append(c)
return "".join(ans)
```

#### Java
Expand All @@ -93,10 +98,9 @@ class Solution {
StringBuilder ans = new StringBuilder();
for (char c : s.toCharArray()) {
int n = ans.length();
if (n > 1 && ans.charAt(n - 1) == c && ans.charAt(n - 2) == c) {
continue;
if (n < 2 || c != ans.charAt(n - 1) || c != ans.charAt(n - 2)) {
ans.append(c);
}
ans.append(c);
}
return ans.toString();
}
Expand All @@ -112,8 +116,9 @@ public:
string ans = "";
for (char& c : s) {
int n = ans.size();
if (n > 1 && ans[n - 1] == c && ans[n - 2] == c) continue;
ans.push_back(c);
if (n < 2 || ans[n - 1] != c || ans[n - 2] != c) {
ans += c;
}
}
return ans;
}
Expand All @@ -126,16 +131,29 @@ public:
func makeFancyString(s string) string {
ans := []rune{}
for _, c := range s {
n := len(ans)
if n > 1 && ans[n-1] == c && ans[n-2] == c {
continue
if n := len(ans); n < 2 || c != ans[n-1] || c != ans[n-2] {
ans = append(ans, c)
}
ans = append(ans, c)
}
return string(ans)
}
```

#### TypeScript

```ts
function makeFancyString(s: string): string {
const ans: string[] = [];
for (const c of s) {
const n = ans.length;
if (n < 2 || c !== ans[n - 1] || c !== ans[n - 2]) {
ans.push(c);
}
}
return ans.join('');
}
```

#### PHP

```php
Expand All @@ -145,15 +163,17 @@ class Solution {
* @return String
*/
function makeFancyString($s) {
$rs = '';
for ($i = 0; $i < strlen($s); $i++) {
if ($s[$i] == $s[$i + 1] && $s[$i] == $s[$i + 2]) {
continue;
} else {
$rs .= $s[$i];
$ans = [];
$length = strlen($s);

for ($i = 0; $i < $length; $i++) {
$n = count($ans);
if ($n < 2 || $s[$i] !== $ans[$n - 1] || $s[$i] !== $ans[$n - 2]) {
$ans[] = $s[$i];
}
}
return $rs;

return implode('', $ans);
}
}
```
Expand Down
7 changes: 4 additions & 3 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -4,9 +4,10 @@ class Solution {
string ans = "";
for (char& c : s) {
int n = ans.size();
if (n > 1 && ans[n - 1] == c && ans[n - 2] == c) continue;
ans.push_back(c);
if (n < 2 || ans[n - 1] != c || ans[n - 2] != c) {
ans += c;
}
}
return ans;
}
};
};
8 changes: 3 additions & 5 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,11 +1,9 @@
func makeFancyString(s string) string {
ans := []rune{}
for _, c := range s {
n := len(ans)
if n > 1 && ans[n-1] == c && ans[n-2] == c {
continue
if n := len(ans); n < 2 || c != ans[n-1] || c != ans[n-2] {
ans = append(ans, c)
}
ans = append(ans, c)
}
return string(ans)
}
}
7 changes: 3 additions & 4 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/Solution.java
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Original file line number Diff line number Diff line change
Expand Up @@ -3,11 +3,10 @@ public String makeFancyString(String s) {
StringBuilder ans = new StringBuilder();
for (char c : s.toCharArray()) {
int n = ans.length();
if (n > 1 && ans.charAt(n - 1) == c && ans.charAt(n - 2) == c) {
continue;
if (n < 2 || c != ans.charAt(n - 1) || c != ans.charAt(n - 2)) {
ans.append(c);
}
ans.append(c);
}
return ans.toString();
}
}
}
16 changes: 9 additions & 7 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/Solution.php
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -4,14 +4,16 @@ class Solution {
* @return String
*/
function makeFancyString($s) {
$rs = '';
for ($i = 0; $i < strlen($s); $i++) {
if ($s[$i] == $s[$i + 1] && $s[$i] == $s[$i + 2]) {
continue;
} else {
$rs .= $s[$i];
$ans = [];
$length = strlen($s);

for ($i = 0; $i < $length; $i++) {
$n = count($ans);
if ($n < 2 || $s[$i] !== $ans[$n - 1] || $s[$i] !== $ans[$n - 2]) {
$ans[] = $s[$i];
}
}
return $rs;

return implode('', $ans);
}
}
7 changes: 3 additions & 4 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2,7 +2,6 @@ class Solution:
def makeFancyString(self, s: str) -> str:
ans = []
for c in s:
if len(ans) > 1 and ans[-1] == ans[-2] == c:
continue
ans.append(c)
return ''.join(ans)
if len(ans) < 2 or ans[-1] != c or ans[-2] != c:
ans.append(c)
return "".join(ans)
10 changes: 10 additions & 0 deletions solution/1900-1999/1957.Delete Characters to Make Fancy String/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
function makeFancyString(s: string): string {
const ans: string[] = [];
for (const c of s) {
const n = ans.length;
if (n < 2 || c !== ans[n - 1] || c !== ans[n - 2]) {
ans.push(c);
}
}
return ans.join('');
}
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