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算法/leetcode/常规题/Q131 分割回文串.md
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算法/leetcode/常规题/Q131 分割回文串.md
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---
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id: "20220309"
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date: "2021/03/09 20:15"
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title: "Q131 分割回文串(palindrome partitioning)"
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tags: ["java", "leetcode", "dp", "dfs"]
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index_img: https://qiniupic.fleyx.com/blog/202203091650691.jpg?imageView2/2/w/200
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banner_img: https://qiniupic.fleyx.com/blog/202203091650691.jpg
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categories:
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- "算法"
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- "leetcode刷题"
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---
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## 解析思路
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leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/problems/palindrome-partitioning/)。
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本题对于从没接触过回溯题目的人来说可以说是非常困难了，并不容易。
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首先题意是要求返回所有的分割组合，使分割后数组中的数都为回文数。
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判断回文数的方法很简单，两个指针从两端往中间遍历即可，或者先用 dp 计算得到所有的回文数组合（本解答采用 dp）,本题字符串最长为 16，暴力判断和 dp 区别不大。
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关键是如何穷举出所有分割可能，简单的 for 循环显然是做不到的.具体分割方法可以先看下图：
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## 字符串分割
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将字符串切分结果展开成一棵树以后，可以发现结果集就是对这棵树进行深度优先搜索的遍历结果,理解清楚分割原理以后用代码实现就比较简单了：
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dfs 分割字符串代码如下：
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```java
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/**
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* 分割字符串
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*
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* @param s 字符串
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* @param index 子串的起始位置
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* @param temp 临时存储已经分割后的字符串
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* @param res 存储最终分割结果
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*
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* @author fanxb
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* date 2022/3/9 16:39
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*/
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private void dfs(String s, int index, Stack<String> temp, List<List<String>> res ) {
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if (index == s.length()) {
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//当字符串起始位置超过字符串长度时说明本条链已经搜索完了
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res.add(new ArrayList<>(temp));
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return;
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}
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for (int i = index; i < s.length(); i++) {
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String tempS = s.substring(index, i + 1);
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temp.push(tempS);
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dfs(s, i + 1, temp, res, dp);
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//这里pop因为此路径已经处理完毕了，pop出去后，进行下一条路径处理
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temp.pop();
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}
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}
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```
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## dp 得到回文数分布情况
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判断回文的递推关系比较容易写.定义 dp[i][j]表示从下标 i 到 j 的字符串是否回文，递推关系如下:
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1. dp[i][j]=true,i==j
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2. dp[i][j]=s[i]==s[j],i=j-1
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3. dp[i][j]=dp[i+1][j-1] && s[i]==s[j],其他情况
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## 代码
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经过上一节，现在我们已经能够穷举字符串所有的分割组合了，结合题意需要留下回文数组合，那么只需要进行剪枝操作即可（判断当前字符串片段是否回文，不回文就没必要继续当前路径）
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完整代码如下：
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```java
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public class Q131 {
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public List<List<String>> partition(String s) {
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//先用dp找到所有符合条件的回文
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boolean[][] dp = new boolean[s.length()][s.length()];
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for (int j = 0; j < dp.length; j++) {
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for (int i = 0; i <= j; i++) {
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if (i == j) {
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dp[i][j] = true;
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} else {
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boolean b = s.charAt(i) == s.charAt(j);
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if (i + 1 == j) {
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dp[i][j] = b;
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} else {
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dp[i][j] = dp[i + 1][j - 1] && b;
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}
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}
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}
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}
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List<List<String>> res = new ArrayList<>();
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dfs(s, 0, new Stack<>(), res, dp);
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return res;
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}
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/**
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* 分割字符串
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*
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* @param s 字符串
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* @param index 子串的起始位置
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* @param temp 临时存储已经分割后的字符串
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* @param res 存储最终分割结果
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* @param dp dp结果
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* @author fanxb
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* date 2022/3/9 16:39
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*/
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private void dfs(String s, int index, Stack<String> temp, List<List<String>> res, boolean[][] dp) {
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if (index == s.length()) {
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res.add(new ArrayList<>(temp));
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return;
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}
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for (int i = index; i < s.length(); i++) {
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if (dp[index][i + 1]) {
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String tempS = s.substring(index, i + 1);
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temp.push(tempS);
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dfs(s, i + 1, temp, res, dp);
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temp.pop();
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}
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}
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}
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}
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```
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算法/leetcode/常规题/Q132 分割回文串Ⅱ.md
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算法/leetcode/常规题/Q132 分割回文串Ⅱ.md
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---
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id: "202203091"
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date: "2021/03/09 22:15"
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title: "Q132 分割回文串2(palindrome-partitioning-ii)"
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tags: ["java", "leetcode", "dp"]
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index_img: https://qiniupic.fleyx.com/blog/202203091652083.png?imageView2/2/w/200
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banner_img: https://qiniupic.fleyx.com/blog/202203091652083.png
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categories:
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- "算法"
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- "leetcode刷题"
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---
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### 解析思路
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leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/problems/palindrome-partitioning-ii/)。
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经过[分割回文数](https://blog.fleyx.com/blog/detail/20220309)后，本题应该就比较简单了。
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题目要求返回最少的分割次数，通常这类最优问题都可以用 dp 来解决。注意这里无法使用上题的 dfs 暴力穷举所有结果后找到最优，会超时(因为字符长度最大为 2000，上题为 16)。
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如果要用 dp 那么就需要构建 dp 表达式：
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定义 minC[k]表示从 0 到 k 的最小分割次数，递推关系如下：
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1. minC[k] = 0 ,k==0 || dp[0][k],当 k=0 或者 0-k 是一个回文串
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2. minC[k] = min(min[r-1]+1).其他情况需要在 0 到 k 之间找到一个 r 使 r 到 k 为一个回文数，那么分割次数就是 minC[r-1]+1(0 到 r-1 需要的最小分割次数再加上一次分割),在所有的情况下取最小的那个即可
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### 代码如下：
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```java
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public class Q132 {
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public int minCut(String s) {
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//先用dp找到所有符合条件的回文
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boolean[][] dp = new boolean[s.length()][s.length()];
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for (int j = 0; j < dp.length; j++) {
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for (int i = 0; i <= j; i++) {
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if (i == j) {
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dp[i][j] = true;
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} else {
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boolean b = s.charAt(i) == s.charAt(j);
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if (i + 1 == j) {
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dp[i][j] = b;
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} else {
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dp[i][j] = dp[i + 1][j - 1] && b;
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}
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}
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}
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}
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//再次dp找到最小分割次数.定义minArr[i]为从0到i的最小分割次数
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int[] minArr = new int[s.length()];
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for (int i = 1; i < s.length(); i++) {
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//定义l，当l->i构成回文串时，此时分割次数为=minArr[l-1]+1.找到最小的l
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int temp = i;
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for (int l = 0; l <= i; l++) {
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if (dp[l][i]) {
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temp = Math.min(temp, l == 0 ? 0 : minArr[l - 1] + 1);
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}
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}
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minArr[i] = temp;
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}
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return minArr[s.length() - 1];
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}
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public static void main(String[] args) {
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System.out.println(new Q132().minCut("aab"));
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}
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}
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```
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算法/leetcode/面试题/xml/q131-字符串切割.drawio
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算法/leetcode/面试题/xml/q131-字符串切割.drawio
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