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feat: add solutions to lc problem: No.0647
No.0647.Palindromic Substrings
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‎solution/0600-0699/0647.Palindromic Substrings/README.md

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<!-- 这里可写通用的实现逻辑 -->
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在 Manacher 算法的计算过程中，`p[i] - 1` 表示以第 `i` 位为中心的最大回文长度，`(p[i] - 1) / 2` **向上取整**即可得到以第 `i` 位为中心的回文串数量
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**方法一：从中心向两侧扩展回文串**
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时间复杂度 $O(n^2)$，其中 $n$ 是字符串 `s` 的长度。
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**方法二：Manacher 算法**
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在 Manacher 算法的计算过程中，$p[i]-1$ 表示以第 $i$ 位为中心的最大回文长度，以第 $i$ 位为中心的回文串数量为 $\left \lceil \frac{p[i]-1}{2} \right \rceil$。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是字符串 `s` 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 countSubstrings(self, s: str) -> int:
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ans, n = 0, len(s)
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for k in range(n * 2 - 1):
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i, j = k // 2, (k + 1) // 2
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while ~i and j < n and s[i] == s[j]:
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ans += 1
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i, j = i - 1, j + 1
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return ans
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```
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```python
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class Solution:
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def countSubstrings(self, s: str) -> int:
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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 int countSubstrings(String s) {
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int ans = 0;
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int n = s.length();
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for (int k = 0; k < n * 2 - 1; ++k) {
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int i = k / 2, j = (k + 1) / 2;
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while (i >= 0 && j < n && s.charAt(i) == s.charAt(j)) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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}
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}
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```
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```java
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class Solution {
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public int countSubstrings(String s) {
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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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int countSubstrings(string s) {
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int ans = 0;
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int n = s.size();
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for (int k = 0; k < n * 2 - 1; ++k) {
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int i = k / 2, j = (k + 1) / 2;
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while (~i && j < n && s[i] == s[j]) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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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 countSubstrings(s string) int {
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ans, n := 0, len(s)
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for k := 0; k < n*2-1; k++ {
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i, j := k/2, (k+1)/2
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for i >= 0 && j < n && s[i] == s[j] {
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ans++
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i, j = i-1, j+1
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}
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}
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return ans
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {string} s
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* @return {number}
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*/
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var countSubstrings = function (s) {
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let ans = 0;
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const n = s.length;
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for (let k = 0; k < n * 2 - 1; ++k) {
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let i = k >> 1;
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let j = (k + 1) >> 1;
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while (~i && j < n && s[i] == s[j]) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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};
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```
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### **...**
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```

‎solution/0600-0699/0647.Palindromic Substrings/README_EN.md

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### **Python3**
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```python
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class Solution:
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def countSubstrings(self, s: str) -> int:
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ans, n = 0, len(s)
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for k in range(n * 2 - 1):
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i, j = k // 2, (k + 1) // 2
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while ~i and j < n and s[i] == s[j]:
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ans += 1
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i, j = i - 1, j + 1
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return ans
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```
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```python
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class Solution:
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def countSubstrings(self, s: str) -> int:
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### **Java**
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```java
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class Solution {
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public int countSubstrings(String s) {
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int ans = 0;
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int n = s.length();
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for (int k = 0; k < n * 2 - 1; ++k) {
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int i = k / 2, j = (k + 1) / 2;
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while (i >= 0 && j < n && s.charAt(i) == s.charAt(j)) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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}
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}
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```
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```java
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class Solution {
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public int countSubstrings(String s) {
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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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int countSubstrings(string s) {
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int ans = 0;
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int n = s.size();
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for (int k = 0; k < n * 2 - 1; ++k) {
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int i = k / 2, j = (k + 1) / 2;
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while (~i && j < n && s[i] == s[j]) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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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 countSubstrings(s string) int {
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ans, n := 0, len(s)
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for k := 0; k < n*2-1; k++ {
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i, j := k/2, (k+1)/2
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for i >= 0 && j < n && s[i] == s[j] {
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ans++
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i, j = i-1, j+1
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}
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}
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return ans
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {string} s
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* @return {number}
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*/
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var countSubstrings = function (s) {
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let ans = 0;
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const n = s.length;
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for (let k = 0; k < n * 2 - 1; ++k) {
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let i = k >> 1;
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let j = (k + 1) >> 1;
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while (~i && j < n && s[i] == s[j]) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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};
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```
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### **...**
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```

‎solution/0600-0699/0647.Palindromic Substrings/Solution.cpp

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class Solution {
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public:
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int countSubstrings(string s) {
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int ans = 0;
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int n = s.size();
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for (int k = 0; k < n * 2 - 1; ++k) {
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int i = k / 2, j = (k + 1) / 2;
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while (~i && j < n && s[i] == s[j]) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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}
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};

‎solution/0600-0699/0647.Palindromic Substrings/Solution.go

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func countSubstrings(s string) int {
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ans, n := 0, len(s)
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for k := 0; k < n*2-1; k++ {
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i, j := k/2, (k+1)/2
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for i >= 0 && j < n && s[i] == s[j] {
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ans++
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i, j = i-1, j+1
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}
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}
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return ans
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}

‎solution/0600-0699/0647.Palindromic Substrings/Solution.js

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/**
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* @param {string} s
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* @return {number}
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*/
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var countSubstrings = function (s) {
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let ans = 0;
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const n = s.length;
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for (let k = 0; k < n * 2 - 1; ++k) {
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let i = k >> 1;
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let j = (k + 1) >> 1;
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while (~i && j < n && s[i] == s[j]) {
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++ans;
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--i;
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++j;
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}
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}
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return ans;
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};

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