feat: azl397985856#5 azl397985856#516

Author: luzhipeng

README.en.md

@@ -125,6 +125,7 @@ The data structures mainly includes:
 
 - [0002. Add Two Numbers](./problems/2.addTwoNumbers.md)
 - [0003. Longest Substring Without Repeating Characters](./problems/3.longestSubstringWithoutRepeatingCharacters.md)
+- [0005.longest-palindromic-substring](./problems/5.longest-palindromic-substring.md) 🆕
 - [0011.container-with-most-water](./problems/11.container-with-most-water.md)
 - [0015.3-sum](./problems/15.3-sum.md)
 - [0019. Remove Nth Node From End of List](./problems/19.removeNthNodeFromEndofList.md)
@@ -173,6 +174,7 @@ The data structures mainly includes:
 - [0445.add-two-numbers-ii](./problems/445.add-two-numbers-ii.md)
 - [0454.4-sum-ii](./problems/454.4-sum-ii.md)
 - [0494.target-sum](./problems/494.target-sum.md)
+- [0516.longest-palindromic-subsequence](./problems/516.longest-palindromic-subsequence.md) 🆕
 - [0518.coin-change-2](./problems/518.coin-change-2.md)
 - [0609.find-duplicate-file-in-system](./problems/609.find-duplicate-file-in-system.md) 🆕
 - [0875.koko-eating-bananas](./problems/875.koko-eating-bananas.md)

README.md

@@ -123,6 +123,7 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 
 - [0002. Add Two Numbers](./problems/2.addTwoNumbers.md)
 - [0003. Longest Substring Without Repeating Characters](./problems/3.longestSubstringWithoutRepeatingCharacters.md)
+- [0005.longest-palindromic-substring](./problems/5.longest-palindromic-substring.md) 🆕
 - [0011.container-with-most-water](./problems/11.container-with-most-water.md)
 - [0015.3-sum](./problems/15.3-sum.md)
 - [0019. Remove Nth Node From End of List](./problems/19.removeNthNodeFromEndofList.md)
@@ -171,6 +172,7 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [0445.add-two-numbers-ii](./problems/445.add-two-numbers-ii.md)
 - [0454.4-sum-ii](./problems/454.4-sum-ii.md)
 - [0494.target-sum](./problems/494.target-sum.md)
+- [0516.longest-palindromic-subsequence](./problems/516.longest-palindromic-subsequence.md) 🆕
 - [0518.coin-change-2](./problems/518.coin-change-2.md)
 - [0609.find-duplicate-file-in-system](./problems/609.find-duplicate-file-in-system.md) 🆕
 - [0875.koko-eating-bananas](./problems/875.koko-eating-bananas.md)

assets/drawio/5.longest-palindromic-substring.drawio

@@ -0,0 +1 @@
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assets/problems/5.longest-palindromic-substring-1.png

@@ -0,0 +1,98 @@
+## 题目地址
+
+https://leetcode.com/problems/longest-palindromic-substring/description/
+
+## 题目描述
+
+```
+Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
+
+Example 1:
+
+Input: "babad"
+Output: "bab"
+Note: "aba" is also a valid answer.
+Example 2:
+
+Input: "cbbd"
+Output: "bb"
+```
+
+## 思路
+
+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串。
+
+![5.longest-palindromic-substring](../assets/problems/5.longest-palindromic-substring-1.png)
+
+解决这类问题的核心思想就是两个字“延伸”，具体来说
+
+- 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
+- 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
+
+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，
+基于此，我们可以建立动态规划模型。
+
+我们可以用 dp[i][j] 表示 s 中从 i 到 j（包括 i 和 j）是否可以形成回文，
+状态转移方程只是将上面的描述转化为代码即可：
+
+```js
+if (s[i] === s[j] && dp[i + 1][j - 1]) {
+  dp[i][j] = true;
+}
+```
+![5.longest-palindromic-substring-2](../assets/problems/5.longest-palindromic-substring-2.png)
+
+base case就是一个字符（轴对称点是本身），或者两个字符（轴对称点是介于两者之间的虚拟点）。
+
+![5.longest-palindromic-substring-3](../assets/problems/5.longest-palindromic-substring-3.png)
+## 关键点
+
+- ”延伸“（extend）
+
+## 代码
+
+```js
+/*
+ * @lc app=leetcode id=5 lang=javascript
+ *
+ * [5] Longest Palindromic Substring
+ */
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var longestPalindrome = function(s) {
+  // babad
+  // tag : dp
+  if (!s || s.length === 0) return "";
+  let res = s[0];
+
+  const dp = [];
+
+  // 倒着遍历简化操作， 这么做的原因是dp[i][..]依赖于dp[i + 1][..]
+  for (let i = s.length - 1; i >= 0; i--) {
+    dp[i] = [];
+    for (let j = i; j < s.length; j++) {
+      if (j - i === 0) dp[i][j] = true;
+      // specail case 1
+      else if (j - i === 1 && s[i] === s[j]) dp[i][j] = true;
+      // specail case 2
+      else if (s[i] === s[j] && dp[i + 1][j - 1]) {
+        // state transition
+        dp[i][j] = true;
+      }
+
+      if (dp[i][j] && j - i + 1 > res.length) {
+        // update res
+        res = s.slice(i, j + 1);
+      }
+    }
+  }
+
+  return res;
+};
+```
+
+## 相关题目
+
+-[516.longest-palindromic-subsequence](./516.longest-palindromic-subsequence.md)

assets/problems/5.longest-palindromic-substring-2.png

@@ -0,0 +1,96 @@
+## 题目地址
+
+https://leetcode.com/problems/longest-palindromic-subsequence/description/
+
+## 题目描述
+
+```
+Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.
+
+Example 1:
+Input:
+
+"bbbab"
+Output:
+4
+One possible longest palindromic subsequence is "bbbb".
+Example 2:
+Input:
+
+"cbbd"
+Output:
+2
+One possible longest palindromic subsequence is "bb".
+```
+
+## 思路
+
+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。
+
+![516.longest-palindromic-subsequence-1](../assets/problems/516.longest-palindromic-subsequence-1.png)
+
+解决这类问题的核心思想就是两个字“延伸”，具体来说
+
+- 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`
+- 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`
+
+![516.longest-palindromic-subsequence-2](../assets/problems/516.longest-palindromic-subsequence-2.png)
+
+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，
+基于此，我们可以建立动态规划模型。
+
+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，
+状态转移方程只是将上面的描述转化为代码即可：
+
+```js
+if (s[i] === s[j]) {
+  dp[i][j] = dp[i + 1][j - 1] + 2;
+} else {
+  dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
+}
+```
+
+base case 就是一个字符（轴对称点是本身）
+
+![516.longest-palindromic-subsequence-3](../assets/problems/516.longest-palindromic-subsequence-3.png)
+
+## 关键点
+
+- ”延伸“（extend）
+
+## 代码
+
+```js
+/*
+ * @lc app=leetcode id=516 lang=javascript
+ *
+ * [516] Longest Palindromic Subsequence
+ */
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var longestPalindromeSubseq = function(s) {
+  // bbbab 返回4
+  // tag : dp
+  const dp = [];
+
+  for (let i = s.length - 1; i >= 0; i--) {
+    dp[i] = Array(s.length).fill(0);
+    for (let j = i; j < s.length; j++) {
+      if (i - j === 0) dp[i][j] = 1;
+      else if (s[i] === s[j]) {
+        dp[i][j] = dp[i + 1][j - 1] + 2;
+      } else {
+        dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
+      }
+    }
+  }
+
+  return dp[0][s.length - 1];
+};
+```
+
+## 相关题目
+
+-[5.longest-palindromic-substring](./5.longest-palindromic-substring.md)

assets/problems/5.longest-palindromic-substring-3.png

@@ -0,0 +1,49 @@
+## 字符串问题
+
+字符串问题有很多，从简单的实现substr，识别回文，到复杂一点的公共子串/子序列。其实字符串本质上也是字符数组，因此
+很多数据的思想和方法也可以用在字符串问题上，并且在有些时候能够发挥很好的作用。
+
+专门处理字符串的算法也很多，比如trie，马拉车算法，游程编码，huffman树等等。
+
+
+## 实现字符串的一些原生方法
+
+这类题目应该是最直接的题目了，题目歧义比较小, 难度也是相对较小，因此用于电面等形式也是不错的。
+
+- [28.implement-str-str](https://leetcode.com/problems/implement-strstr/)
+- [344.reverse-string](../backlog/344.reverse-string.js)
+
+## 回文
+
+回文串就是一个正读和反读都一样的字符串，比如“level”或者“noon”等等就是回文串。
+
+判断是否回文的通用方法是首尾双指针，具体可以见下方125号题目。 判断最长回文的思路主要是两个字"扩展"，
+如果可以充分利用回文的特点，则可以减少很多无谓的计算，典型的是《马拉车算法》。
+
+
+### 相关问题
+
+- [5.longest-palindromic-substring](../problems/5.longest-palindromic-substring.md)
+
+- [125.valid-palindrome](../problems/125.valid-palindrome.md)
+
+- [131.palindrome-partitioning](../problems/131.palindrome-partitioning.md)
+
+- [shortest-palindrome](https://leetcode.com/problems/shortest-palindrome/)
+
+- [516.longest-palindromic-subsequence](../problems/516.longest-palindromic-subsequence.md)
+
+
+## 前缀问题
+
+前缀树用来处理这种问题是最符合直觉的，但是它也有缺点，比如公共前缀很少的情况下，比较费内存。
+
+### 相关题目
+
+-[14.longest-common-prefix](../14.longest-common-prefix.js)
+-[208.implement-trie-prefix-tree](../problems/208.implement-trie-prefix-tree.md)
+
+
+## 其他问题
+
+- [139.word-break](../problems/139.word-break.md)