97 lines
2.5 KiB
Markdown
97 lines
2.5 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/longest-palindromic-subsequence/description/
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## 题目描述
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```
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Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.
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Example 1:
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Input:
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"bbbab"
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Output:
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4
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One possible longest palindromic subsequence is "bbbb".
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Example 2:
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Input:
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"cbbd"
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Output:
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2
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One possible longest palindromic subsequence is "bb".
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```
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## 思路
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这是一道最长回文的题目，要我们求出给定字符串的最大回文子序列。
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解决这类问题的核心思想就是两个字“延伸”，具体来说
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- 如果一个字符串是回文串，那么在它左右分别加上一个相同的字符，那么它一定还是一个回文串，因此`回文长度增加2`
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- 如果一个字符串不是回文串，或者在回文串左右分别加不同的字符，得到的一定不是回文串,因此`回文长度不变，我们取[i][j-1]和[i+1][j]的较大值`
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事实上，上面的分析已经建立了大问题和小问题之间的关联，
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基于此，我们可以建立动态规划模型。
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我们可以用 dp[i][j] 表示 s 中从 i 到 j（包括 i 和 j）的回文序列长度，
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状态转移方程只是将上面的描述转化为代码即可：
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```js
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if (s[i] === s[j]) {
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dp[i][j] = dp[i + 1][j - 1] + 2;
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} else {
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dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
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}
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```
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base case 就是一个字符（轴对称点是本身）
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## 关键点
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- ”延伸“（extend）
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## 代码
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```js
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/*
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* @lc app=leetcode id=516 lang=javascript
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*
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* [516] Longest Palindromic Subsequence
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*/
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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 longestPalindromeSubseq = function(s) {
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// bbbab 返回4
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// tag : dp
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const dp = [];
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for (let i = s.length - 1; i >= 0; i--) {
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dp[i] = Array(s.length).fill(0);
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for (let j = i; j < s.length; j++) {
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if (i - j === 0) dp[i][j] = 1;
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else if (s[i] === s[j]) {
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dp[i][j] = dp[i + 1][j - 1] + 2;
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} else {
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dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
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}
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}
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}
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return dp[0][s.length - 1];
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};
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```
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## 相关题目
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- [5.longest-palindromic-substring](./5.longest-palindromic-substring.md)
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