71 lines
2.6 KiB
Markdown
71 lines
2.6 KiB
Markdown
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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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