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feat: add solutions to lc problem: No.1328 (doocs#1278)
No.1328.Break a Palindrome
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‎solution/1300-1399/1328.Break a Palindrome/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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我们先判断字符串的长度是否为 $1$，若是则直接返回空串。
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否则，我们从左到右遍历字符串的前半部分，找到第一个不为 `'a'` 的字符，将其改为 `'a'` 即可。如果不存在这样的字符，那么我们将最后一个字符改为 `'b'` 即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def breakPalindrome(self, palindrome: str) -> str:
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n = len(palindrome)
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if n == 1:
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return ""
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s = list(palindrome)
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i = 0
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while i < n // 2 and s[i] == "a":
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i += 1
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if i == n // 2:
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s[-1] = "b"
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else:
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s[i] = "a"
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return "".join(s)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String breakPalindrome(String palindrome) {
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int n = palindrome.length();
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if (n == 1) {
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return "";
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}
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char[] cs = palindrome.toCharArray();
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int i = 0;
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while (i < n / 2 && cs[i] == 'a') {
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++i;
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}
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if (i == n / 2) {
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cs[n - 1] = 'b';
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} else {
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cs[i] = 'a';
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}
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return String.valueOf(cs);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string breakPalindrome(string palindrome) {
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int n = palindrome.size();
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if (n == 1) {
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return "";
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}
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int i = 0;
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while (i < n / 2 && palindrome[i] == 'a') {
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++i;
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}
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if (i == n / 2) {
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palindrome[n - 1] = 'b';
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} else {
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palindrome[i] = 'a';
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}
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return palindrome;
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}
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};
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```
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### **Go**
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```go
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func breakPalindrome(palindrome string) string {
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n := len(palindrome)
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if n == 1 {
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return ""
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}
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i := 0
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s := []byte(palindrome)
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for i < n/2 && s[i] == 'a' {
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i++
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}
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if i == n/2 {
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s[n-1] = 'b'
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} else {
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s[i] = 'a'
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}
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return string(s)
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}
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```
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### **TypeScript**
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```ts
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function breakPalindrome(palindrome: string): string {
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const n = palindrome.length;
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if (n === 1) {
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return '';
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}
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const s = palindrome.split('');
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let i = 0;
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while (i < n >> 1 && s[i] === 'a') {
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i++;
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}
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if (i == n >> 1) {
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s[n - 1] = 'b';
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} else {
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s[i] = 'a';
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}
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return s.join('');
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}
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```
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### **...**

‎solution/1300-1399/1328.Break a Palindrome/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -43,13 +43,112 @@ Of all the ways, &quot;aaccba&quot; is the lexicographically smallest.
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### **Python3**
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```python
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class Solution:
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def breakPalindrome(self, palindrome: str) -> str:
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n = len(palindrome)
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if n == 1:
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return ""
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s = list(palindrome)
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i = 0
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while i < n // 2 and s[i] == "a":
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i += 1
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if i == n // 2:
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s[-1] = "b"
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else:
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s[i] = "a"
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return "".join(s)
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```
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### **Java**
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```java
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class Solution {
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public String breakPalindrome(String palindrome) {
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int n = palindrome.length();
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if (n == 1) {
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return "";
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}
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char[] cs = palindrome.toCharArray();
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int i = 0;
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while (i < n / 2 && cs[i] == 'a') {
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++i;
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}
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if (i == n / 2) {
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cs[n - 1] = 'b';
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} else {
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cs[i] = 'a';
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}
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return String.valueOf(cs);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string breakPalindrome(string palindrome) {
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int n = palindrome.size();
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if (n == 1) {
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return "";
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}
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int i = 0;
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while (i < n / 2 && palindrome[i] == 'a') {
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++i;
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}
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if (i == n / 2) {
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palindrome[n - 1] = 'b';
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} else {
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palindrome[i] = 'a';
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}
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return palindrome;
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}
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};
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```
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### **Go**
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```go
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func breakPalindrome(palindrome string) string {
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n := len(palindrome)
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if n == 1 {
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return ""
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}
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i := 0
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s := []byte(palindrome)
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for i < n/2 && s[i] == 'a' {
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i++
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}
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if i == n/2 {
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s[n-1] = 'b'
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} else {
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s[i] = 'a'
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}
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return string(s)
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}
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```
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### **TypeScript**
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```ts
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function breakPalindrome(palindrome: string): string {
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const n = palindrome.length;
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if (n === 1) {
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return '';
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}
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const s = palindrome.split('');
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let i = 0;
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while (i < n >> 1 && s[i] === 'a') {
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i++;
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}
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if (i == n >> 1) {
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s[n - 1] = 'b';
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} else {
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s[i] = 'a';
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}
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return s.join('');
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}
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```
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### **...**

‎solution/1300-1399/1328.Break a Palindrome/Solution.cpp

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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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string breakPalindrome(string palindrome) {
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int n = palindrome.size();
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if (n == 1) {
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return "";
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}
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int i = 0;
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while (i < n / 2 && palindrome[i] == 'a') {
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++i;
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}
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if (i == n / 2) {
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palindrome[n - 1] = 'b';
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} else {
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palindrome[i] = 'a';
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}
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return palindrome;
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}
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};

‎solution/1300-1399/1328.Break a Palindrome/Solution.go

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func breakPalindrome(palindrome string) string {
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n := len(palindrome)
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if n == 1 {
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return ""
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}
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i := 0
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s := []byte(palindrome)
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for i < n/2 && s[i] == 'a' {
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i++
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}
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if i == n/2 {
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s[n-1] = 'b'
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} else {
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s[i] = 'a'
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}
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return string(s)
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}

‎solution/1300-1399/1328.Break a Palindrome/Solution.java

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class Solution {
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public String breakPalindrome(String palindrome) {
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int n = palindrome.length();
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if (n == 1) {
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return "";
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}
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char[] cs = palindrome.toCharArray();
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int i = 0;
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while (i < n / 2 && cs[i] == 'a') {
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++i;
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}
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if (i == n / 2) {
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cs[n - 1] = 'b';
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} else {
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cs[i] = 'a';
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}
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return String.valueOf(cs);
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}
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}

‎solution/1300-1399/1328.Break a Palindrome/Solution.py

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Original file line numberDiff line numberDiff line change
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class Solution:
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def breakPalindrome(self, palindrome: str) -> str:
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n = len(palindrome)
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if n == 1:
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return ""
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s = list(palindrome)
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i = 0
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while i < n // 2 and s[i] == "a":
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i += 1
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if i == n // 2:
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s[-1] = "b"
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else:
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s[i] = "a"
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return "".join(s)

‎solution/1300-1399/1328.Break a Palindrome/Solution.ts

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function breakPalindrome(palindrome: string): string {
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const n = palindrome.length;
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if (n === 1) {
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return '';
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}
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const s = palindrome.split('');
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let i = 0;
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while (i < n >> 1 && s[i] === 'a') {
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i++;
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}
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if (i == n >> 1) {
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s[n - 1] = 'b';
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} else {
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s[i] = 'a';
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}
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return s.join('');
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}

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