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feat: add solutions to lc problem: No.1842 (doocs#2127)
No.1842.Next Palindrome Using Same Digits
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‎solution/1800-1899/1842.Next Palindrome Using Same Digits/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：求前一半的下一个排列**
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根据题目描述，我们只需要求出前一半的下一个排列，然后遍历前一半，对称赋值后半部分即可。
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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 nextPalindrome(self, num: str) -> str:
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def next_permutation(nums: List[str]) -> bool:
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n = len(nums) // 2
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i = n - 2
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while i >= 0 and nums[i] >= nums[i + 1]:
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i -= 1
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if i < 0:
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return False
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j = n - 1
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while j >= 0 and nums[j] <= nums[i]:
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j -= 1
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nums[i], nums[j] = nums[j], nums[i]
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nums[i + 1 : n] = nums[i + 1 : n][::-1]
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return True
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nums = list(num)
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if not next_permutation(nums):
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return ""
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n = len(nums)
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for i in range(n // 2):
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nums[n - i - 1] = nums[i]
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return "".join(nums)
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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 nextPalindrome(String num) {
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char[] nums = num.toCharArray();
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if (!nextPermutation(nums)) {
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return "";
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}
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int n = nums.length;
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for (int i = 0; i < n / 2; ++i) {
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nums[n - 1 - i] = nums[i];
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}
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return String.valueOf(nums);
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}
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private boolean nextPermutation(char[] nums) {
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int n = nums.length / 2;
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int i = n - 2;
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while (i >= 0 && nums[i] >= nums[i + 1]) {
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--i;
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}
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if (i < 0) {
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return false;
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}
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int j = n - 1;
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while (j >= 0 && nums[i] >= nums[j]) {
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--j;
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}
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swap(nums, i++, j);
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for (j = n - 1; i < j; ++i, --j) {
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swap(nums, i, j);
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}
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return true;
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}
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private void swap(char[] nums, int i, int j) {
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char t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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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 nextPalindrome(string num) {
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int n = num.size();
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string nums = num.substr(0, n / 2);
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if (!next_permutation(begin(nums), end(nums))) {
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return "";
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}
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for (int i = 0; i < n / 2; ++i) {
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num[i] = nums[i];
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num[n - i - 1] = nums[i];
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}
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return num;
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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 nextPalindrome(num string) string {
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nums := []byte(num)
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n := len(nums)
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if !nextPermutation(nums) {
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return ""
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}
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for i := 0; i < n/2; i++ {
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nums[n-1-i] = nums[i]
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}
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return string(nums)
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}
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func nextPermutation(nums []byte) bool {
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n := len(nums) / 2
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i := n - 2
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for i >= 0 && nums[i] >= nums[i+1] {
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i--
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}
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if i < 0 {
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return false
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}
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j := n - 1
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for j >= 0 && nums[j] <= nums[i] {
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j--
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}
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nums[i], nums[j] = nums[j], nums[i]
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for i, j = i+1, n-1; i < j; i, j = i+1, j-1 {
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nums[i], nums[j] = nums[j], nums[i]
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}
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return true
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}
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```
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### **TypeScript**
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```ts
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function nextPalindrome(num: string): string {
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const nums = num.split('');
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const n = nums.length;
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if (!nextPermutation(nums)) {
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return '';
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}
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for (let i = 0; i < n >> 1; ++i) {
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nums[n - 1 - i] = nums[i];
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}
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return nums.join('');
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}
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function nextPermutation(nums: string[]): boolean {
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const n = nums.length >> 1;
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let i = n - 2;
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while (i >= 0 && nums[i] >= nums[i + 1]) {
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i--;
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}
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if (i < 0) {
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return false;
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}
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let j = n - 1;
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while (j >= 0 && nums[i] >= nums[j]) {
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j--;
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}
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[nums[i], nums[j]] = [nums[j], nums[i]];
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for (i = i + 1, j = n - 1; i < j; ++i, --j) {
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[nums[i], nums[j]] = [nums[j], nums[i]];
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}
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return true;
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}
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```
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### **...**

‎solution/1800-1899/1842.Next Palindrome Using Same Digits/README_EN.md

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## Solutions
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**Solution 1: Find the Next Permutation of the First Half**
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According to the problem description, we only need to find the next permutation of the first half of the string, then traverse the first half and symmetrically assign values to the second half.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def nextPalindrome(self, num: str) -> str:
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def next_permutation(nums: List[str]) -> bool:
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n = len(nums) // 2
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i = n - 2
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while i >= 0 and nums[i] >= nums[i + 1]:
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i -= 1
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if i < 0:
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return False
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j = n - 1
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while j >= 0 and nums[j] <= nums[i]:
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j -= 1
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nums[i], nums[j] = nums[j], nums[i]
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nums[i + 1 : n] = nums[i + 1 : n][::-1]
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return True
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nums = list(num)
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if not next_permutation(nums):
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return ""
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n = len(nums)
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for i in range(n // 2):
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nums[n - i - 1] = nums[i]
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return "".join(nums)
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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 nextPalindrome(String num) {
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char[] nums = num.toCharArray();
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if (!nextPermutation(nums)) {
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return "";
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}
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int n = nums.length;
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for (int i = 0; i < n / 2; ++i) {
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nums[n - 1 - i] = nums[i];
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}
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return String.valueOf(nums);
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}
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private boolean nextPermutation(char[] nums) {
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int n = nums.length / 2;
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int i = n - 2;
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while (i >= 0 && nums[i] >= nums[i + 1]) {
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--i;
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}
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if (i < 0) {
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return false;
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}
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int j = n - 1;
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while (j >= 0 && nums[i] >= nums[j]) {
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--j;
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}
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swap(nums, i++, j);
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for (j = n - 1; i < j; ++i, --j) {
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swap(nums, i, j);
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}
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return true;
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}
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private void swap(char[] nums, int i, int j) {
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char t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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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 nextPalindrome(string num) {
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int n = num.size();
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string nums = num.substr(0, n / 2);
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if (!next_permutation(begin(nums), end(nums))) {
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return "";
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}
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for (int i = 0; i < n / 2; ++i) {
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num[i] = nums[i];
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num[n - i - 1] = nums[i];
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}
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return num;
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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 nextPalindrome(num string) string {
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nums := []byte(num)
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n := len(nums)
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if !nextPermutation(nums) {
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return ""
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}
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for i := 0; i < n/2; i++ {
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nums[n-1-i] = nums[i]
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}
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return string(nums)
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}
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func nextPermutation(nums []byte) bool {
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n := len(nums) / 2
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i := n - 2
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for i >= 0 && nums[i] >= nums[i+1] {
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i--
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}
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if i < 0 {
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return false
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}
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j := n - 1
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for j >= 0 && nums[j] <= nums[i] {
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j--
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}
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nums[i], nums[j] = nums[j], nums[i]
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for i, j = i+1, n-1; i < j; i, j = i+1, j-1 {
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nums[i], nums[j] = nums[j], nums[i]
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}
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return true
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}
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```
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### **TypeScript**
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```ts
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function nextPalindrome(num: string): string {
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const nums = num.split('');
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const n = nums.length;
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if (!nextPermutation(nums)) {
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return '';
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}
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for (let i = 0; i < n >> 1; ++i) {
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nums[n - 1 - i] = nums[i];
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}
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return nums.join('');
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}
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function nextPermutation(nums: string[]): boolean {
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const n = nums.length >> 1;
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let i = n - 2;
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while (i >= 0 && nums[i] >= nums[i + 1]) {
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i--;
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}
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if (i < 0) {
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return false;
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}
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let j = n - 1;
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while (j >= 0 && nums[i] >= nums[j]) {
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j--;
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}
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[nums[i], nums[j]] = [nums[j], nums[i]];
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for (i = i + 1, j = n - 1; i < j; ++i, --j) {
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[nums[i], nums[j]] = [nums[j], nums[i]];
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}
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return true;
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}
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```
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### **...**

‎solution/1800-1899/1842.Next Palindrome Using Same Digits/Solution.cpp

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class Solution {
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public:
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string nextPalindrome(string num) {
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int n = num.size();
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string nums = num.substr(0, n / 2);
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if (!next_permutation(begin(nums), end(nums))) {
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return "";
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}
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for (int i = 0; i < n / 2; ++i) {
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num[i] = nums[i];
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num[n - i - 1] = nums[i];
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}
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return num;
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}
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};

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