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Dec 19, 2023
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feat: add solutions to lc problem: No.1842 #2127

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160 changes: 159 additions & 1 deletion solution/1800-1899/1842.Next Palindrome Using Same Digits/README.md
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Expand Up @@ -51,22 +51,180 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：求前一半的下一个排列**

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。

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

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def nextPalindrome(self, num: str) -> str:
def next_permutation(nums: List[str]) -> bool:
n = len(nums) // 2
i = n - 2
while i >= 0 and nums[i] >= nums[i + 1]:
i -= 1
if i < 0:
return False
j = n - 1
while j >= 0 and nums[j] <= nums[i]:
j -= 1
nums[i], nums[j] = nums[j], nums[i]
nums[i + 1 : n] = nums[i + 1 : n][::-1]
return True

nums = list(num)
if not next_permutation(nums):
return ""
n = len(nums)
for i in range(n // 2):
nums[n - i - 1] = nums[i]
return "".join(nums)
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public String nextPalindrome(String num) {
char[] nums = num.toCharArray();
if (!nextPermutation(nums)) {
return "";
}
int n = nums.length;
for (int i = 0; i < n / 2; ++i) {
nums[n - 1 - i] = nums[i];
}
return String.valueOf(nums);
}

private boolean nextPermutation(char[] nums) {
int n = nums.length / 2;
int i = n - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
--i;
}
if (i < 0) {
return false;
}
int j = n - 1;
while (j >= 0 && nums[i] >= nums[j]) {
--j;
}
swap(nums, i++, j);
for (j = n - 1; i < j; ++i, --j) {
swap(nums, i, j);
}
return true;
}

private void swap(char[] nums, int i, int j) {
char t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
```

### **C++**

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

### **Go**

```go
func nextPalindrome(num string) string {
nums := []byte(num)
n := len(nums)
if !nextPermutation(nums) {
return ""
}
for i := 0; i < n/2; i++ {
nums[n-1-i] = nums[i]
}
return string(nums)
}

func nextPermutation(nums []byte) bool {
n := len(nums) / 2
i := n - 2
for i >= 0 && nums[i] >= nums[i+1] {
i--
}
if i < 0 {
return false
}
j := n - 1
for j >= 0 && nums[j] <= nums[i] {
j--
}
nums[i], nums[j] = nums[j], nums[i]
for i, j = i+1, n-1; i < j; i, j = i+1, j-1 {
nums[i], nums[j] = nums[j], nums[i]
}
return true
}
```

### **TypeScript**

```ts
function nextPalindrome(num: string): string {
const nums = num.split('');
const n = nums.length;
if (!nextPermutation(nums)) {
return '';
}
for (let i = 0; i < n >> 1; ++i) {
nums[n - 1 - i] = nums[i];
}
return nums.join('');
}

function nextPermutation(nums: string[]): boolean {
const n = nums.length >> 1;
let i = n - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
i--;
}
if (i < 0) {
return false;
}
let j = n - 1;
while (j >= 0 && nums[i] >= nums[j]) {
j--;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
for (i = i + 1, j = n - 1; i < j; ++i, --j) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}
return true;
}
```

### **...**
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160 changes: 159 additions & 1 deletion solution/1800-1899/1842.Next Palindrome Using Same Digits/README_EN.md
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## Solutions

**Solution 1: Find the Next Permutation of the First Half**

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.

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

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def nextPalindrome(self, num: str) -> str:
def next_permutation(nums: List[str]) -> bool:
n = len(nums) // 2
i = n - 2
while i >= 0 and nums[i] >= nums[i + 1]:
i -= 1
if i < 0:
return False
j = n - 1
while j >= 0 and nums[j] <= nums[i]:
j -= 1
nums[i], nums[j] = nums[j], nums[i]
nums[i + 1 : n] = nums[i + 1 : n][::-1]
return True

nums = list(num)
if not next_permutation(nums):
return ""
n = len(nums)
for i in range(n // 2):
nums[n - i - 1] = nums[i]
return "".join(nums)
```

### **Java**

```java
class Solution {
public String nextPalindrome(String num) {
char[] nums = num.toCharArray();
if (!nextPermutation(nums)) {
return "";
}
int n = nums.length;
for (int i = 0; i < n / 2; ++i) {
nums[n - 1 - i] = nums[i];
}
return String.valueOf(nums);
}

private boolean nextPermutation(char[] nums) {
int n = nums.length / 2;
int i = n - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
--i;
}
if (i < 0) {
return false;
}
int j = n - 1;
while (j >= 0 && nums[i] >= nums[j]) {
--j;
}
swap(nums, i++, j);
for (j = n - 1; i < j; ++i, --j) {
swap(nums, i, j);
}
return true;
}

private void swap(char[] nums, int i, int j) {
char t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
```

### **C++**

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

### **Go**

```go
func nextPalindrome(num string) string {
nums := []byte(num)
n := len(nums)
if !nextPermutation(nums) {
return ""
}
for i := 0; i < n/2; i++ {
nums[n-1-i] = nums[i]
}
return string(nums)
}

func nextPermutation(nums []byte) bool {
n := len(nums) / 2
i := n - 2
for i >= 0 && nums[i] >= nums[i+1] {
i--
}
if i < 0 {
return false
}
j := n - 1
for j >= 0 && nums[j] <= nums[i] {
j--
}
nums[i], nums[j] = nums[j], nums[i]
for i, j = i+1, n-1; i < j; i, j = i+1, j-1 {
nums[i], nums[j] = nums[j], nums[i]
}
return true
}
```

### **TypeScript**

```ts
function nextPalindrome(num: string): string {
const nums = num.split('');
const n = nums.length;
if (!nextPermutation(nums)) {
return '';
}
for (let i = 0; i < n >> 1; ++i) {
nums[n - 1 - i] = nums[i];
}
return nums.join('');
}

function nextPermutation(nums: string[]): boolean {
const n = nums.length >> 1;
let i = n - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
i--;
}
if (i < 0) {
return false;
}
let j = n - 1;
while (j >= 0 && nums[i] >= nums[j]) {
j--;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
for (i = i + 1, j = n - 1; i < j; ++i, --j) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}
return true;
}
```

### **...**
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15 changes: 15 additions & 0 deletions solution/1800-1899/1842.Next Palindrome Using Same Digits/Solution.cpp
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@@ -0,0 +1,15 @@
class Solution {
public:
string nextPalindrome(string num) {
int n = num.size();
string nums = num.substr(0, n / 2);
if (!next_permutation(begin(nums), end(nums))) {
return "";
}
for (int i = 0; i < n / 2; ++i) {
num[i] = nums[i];
num[n - i - 1] = nums[i];
}
return num;
}
};
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