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feat: add solutions to lc/lcof2 problems
lc No.0131, lcof2 No.086: Palindrome Partitioning
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‎lcof2/剑指 Offer II 086. 分割回文子字符串/README.md

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<p><meta charset="UTF-8" />注意：本题与主站 131&nbsp;题相同：&nbsp;<a href="https://leetcode-cn.com/problems/palindrome-partitioning/">https://leetcode-cn.com/problems/palindrome-partitioning/</a></p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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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 partition(self, s: str) -> List[List[str]]:
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ans = []
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n = len(s)
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dp = [[True] * n for _ in range(n)]
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for i in range(n - 1, -1, -1):
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for j in range(i + 1, n):
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dp[i][j] = s[i] == s[j] and dp[i + 1][j - 1]
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def dfs(s, i, t):
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nonlocal n
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if i == n:
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ans.append(t.copy())
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return
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for j in range(i, n):
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if dp[i][j]:
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t.append(s[i: j + 1])
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dfs(s, j + 1, t)
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t.pop(-1)
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dfs(s, 0, [])
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return ans
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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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private boolean[][] dp;
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private List<List<String>> ans;
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private int n;
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public String[][] partition(String s) {
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ans = new ArrayList<>();
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n = s.length();
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dp = new boolean[n][n];
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for (int i = 0; i < n; ++i) {
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Arrays.fill(dp[i], true);
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}
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for (int i = n - 1; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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dp[i][j] = s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1];
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}
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}
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dfs(s, 0, new ArrayList<>());
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String [][] res = new String [ans.size()][];
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for (int i = 0; i < ans.size(); ++i) {
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res[i] = ans.get(i).toArray(new String[0]);
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}
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return res;
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}
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private void dfs(String s, int i, List<String> t) {
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if (i == n) {
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ans.add(new ArrayList<>(t));
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return;
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}
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for (int j = i; j < n; ++j) {
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if (dp[i][j]) {
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t.add(s.substring(i, j + 1));
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dfs(s, j + 1, t);
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t.remove(t.size() - 1);
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}
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}
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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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vector<vector<bool>> dp;
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vector<vector<string>> ans;
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int n;
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vector<vector<string>> partition(string s) {
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n = s.size();
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dp.assign(n, vector<bool>(n, true));
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for (int i = n - 1; i >= 0; --i)
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{
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for (int j = i + 1; j < n; ++j)
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{
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dp[i][j] = s[i] == s[j] && dp[i + 1][j - 1];
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}
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}
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vector<string> t;
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dfs(s, 0, t);
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return ans;
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}
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void dfs(string& s, int i, vector<string> t) {
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if (i == n)
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{
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ans.push_back(t);
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return;
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}
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for (int j = i; j < n; ++j)
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{
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if (dp[i][j])
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{
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t.push_back(s.substr(i, j - i + 1));
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dfs(s, j + 1, t);
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t.pop_back();
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}
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}
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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 partition(s string) [][]string {
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n := len(s)
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dp := make([][]bool, n)
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var ans [][]string
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for i := 0; i < n; i++ {
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dp[i] = make([]bool, n)
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for j := 0; j < n; j++ {
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dp[i][j] = true
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}
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}
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for i := n - 1; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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dp[i][j] = s[i] == s[j] && dp[i+1][j-1]
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}
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}
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var dfs func(s string, i int, t []string)
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dfs = func(s string, i int, t []string) {
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if i == n {
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ans = append(ans, append([]string(nil), t...))
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return
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}
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for j := i; j < n; j++ {
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if dp[i][j] {
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t = append(t, s[i:j+1])
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dfs(s, j+1, t)
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t = t[:len(t)-1]
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}
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}
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}
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var t []string
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dfs(s, 0, t)
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return ans
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}
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```
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### **...**

‎lcof2/剑指 Offer II 086. 分割回文子字符串/Solution.cpp

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class Solution {
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public:
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vector<vector<bool>> dp;
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vector<vector<string>> ans;
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int n;
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vector<vector<string>> partition(string s) {
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n = s.size();
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dp.assign(n, vector<bool>(n, true));
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for (int i = n - 1; i >= 0; --i)
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{
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for (int j = i + 1; j < n; ++j)
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{
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dp[i][j] = s[i] == s[j] && dp[i + 1][j - 1];
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}
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}
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vector<string> t;
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dfs(s, 0, t);
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return ans;
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}
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void dfs(string& s, int i, vector<string> t) {
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if (i == n)
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{
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ans.push_back(t);
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return;
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}
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for (int j = i; j < n; ++j)
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{
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if (dp[i][j])
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{
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t.push_back(s.substr(i, j - i + 1));
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dfs(s, j + 1, t);
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t.pop_back();
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}
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}
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}
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};

‎lcof2/剑指 Offer II 086. 分割回文子字符串/Solution.go

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func partition(s string) [][]string {
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n := len(s)
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dp := make([][]bool, n)
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var ans [][]string
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for i := 0; i < n; i++ {
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dp[i] = make([]bool, n)
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for j := 0; j < n; j++ {
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dp[i][j] = true
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}
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}
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for i := n - 1; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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dp[i][j] = s[i] == s[j] && dp[i+1][j-1]
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}
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}
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var dfs func(s string, i int, t []string)
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dfs = func(s string, i int, t []string) {
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if i == n {
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ans = append(ans, append([]string(nil), t...))
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return
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}
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for j := i; j < n; j++ {
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if dp[i][j] {
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t = append(t, s[i:j+1])
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dfs(s, j+1, t)
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t = t[:len(t)-1]
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}
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}
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}
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var t []string
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dfs(s, 0, t)
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return ans
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}

‎lcof2/剑指 Offer II 086. 分割回文子字符串/Solution.java

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class Solution {
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private boolean[][] dp;
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private List<List<String>> ans;
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private int n;
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public String[][] partition(String s) {
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ans = new ArrayList<>();
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n = s.length();
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dp = new boolean[n][n];
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for (int i = 0; i < n; ++i) {
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Arrays.fill(dp[i], true);
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}
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for (int i = n - 1; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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dp[i][j] = s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1];
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}
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}
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dfs(s, 0, new ArrayList<>());
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String [][] res = new String [ans.size()][];
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for (int i = 0; i < ans.size(); ++i) {
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res[i] = ans.get(i).toArray(new String[0]);
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}
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return res;
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}
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private void dfs(String s, int i, List<String> t) {
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if (i == n) {
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ans.add(new ArrayList<>(t));
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return;
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}
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for (int j = i; j < n; ++j) {
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if (dp[i][j]) {
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t.add(s.substring(i, j + 1));
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dfs(s, j + 1, t);
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t.remove(t.size() - 1);
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}
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}
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}
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}

‎lcof2/剑指 Offer II 086. 分割回文子字符串/Solution.py

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class Solution:
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def partition(self, s: str) -> List[List[str]]:
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ans = []
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n = len(s)
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dp = [[True] * n for _ in range(n)]
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for i in range(n - 1, -1, -1):
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for j in range(i + 1, n):
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dp[i][j] = s[i] == s[j] and dp[i + 1][j - 1]
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def dfs(s, i, t):
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nonlocal n
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if i == n:
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ans.append(t.copy())
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return
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for j in range(i, n):
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if dp[i][j]:
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t.append(s[i: j + 1])
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dfs(s, j + 1, t)
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t.pop(-1)
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dfs(s, 0, [])
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return ans

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