三刷5

Author: diguage

docs/0005-longest-palindromic-substring.adoc

@@ -1,26 +1,34 @@
 [#0005-longest-palindromic-substring]
-= 5. Longest Palindromic Substring
+= 5. 最长回文子串
 
-{leetcode}/problems/longest-palindromic-substring/[LeetCode - Longest Palindromic Substring^]
+https://leetcode.cn/problems/longest-palindromic-substring/[LeetCode - 5. 最长回文子串 ^]
 
-Given a string *s*, find the longest palindromic substring in *s*. You may assume that the maximum length of *s* is 1000.
+给你一个字符串 `+s+`，找到 `+s+` 中最长的 回文 子串。
 
-.Example 1:
-[subs="verbatim,quotes,macros"]
-----
-*Input:* "babad"
-*Output:* "bab"
-*Note:* "aba" is also a valid answer.
-----
+*示例 1：*
+
+....
+输入：s = "babad"
+输出："bab"
+解释："aba" 同样是符合题意的答案。
+....
+
+*示例 2：*
+
+....
+输入：s = "cbbd"
+输出："bb"
+....
+
+
+*提示：*
+
+* `+1 <= s.length <= 1000+`
+* `+s+` 仅由数字和英文字母组成
 
-.Example 2:
-[subs="verbatim,quotes,macros"]
-----
-*Input:* "cbbd"
-*Output:* "bb"
-----
 
-== 解题分析
+
+== 思路分析
 
 最简单的方式：使用两个指针，把字符串逐个"拆成"子串，然后留下最大子串即可。可惜，这种算法的时间复杂度是 O(n^3^)。
 
@@ -32,7 +40,6 @@ image::images/0005-01.png[]
 
 另外，这道题还可以练手动态规划。
 
-
 [[src-0005]]
 [tabs]
 ====
@@ -53,8 +60,18 @@ include::{sourcedir}/_0005_LongestPalindromicSubstring.java[tag=answer]
 include::{sourcedir}/_0005_LongestPalindromicSubstring_2.java[tag=answer]
 ----
 --
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0005_LongestPalindromicSubstring_3.java[tag=answer]
+----
+--
 ====
 
+
 == 参考资料
 
 * https://www.geeksforgeeks.org/manachers-algorithm-linear-time-longest-palindromic-substring-part-1/[Manacher's Algorithm - Linear Time Longest Palindromic Substring - Part 1^]
@@ -69,5 +86,6 @@ include::{sourcedir}/_0005_LongestPalindromicSubstring_2.java[tag=answer]
 * https://leetcode.cn/problems/longest-palindromic-substring/solutions/9001/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-bao-gu/[5. 最长回文子串 - 详细通俗的思路分析，多解法^]
 * https://leetcode.cn/problems/longest-palindromic-substring/solutions/4299/zui-chang-hui-wen-zi-chuan-c-by-gpe3dbjds1/[5. 最长回文子串^]
 * https://leetcode.cn/problems/longest-palindromic-substring/solutions/7792/zhong-xin-kuo-san-dong-tai-gui-hua-by-liweiwei1419/[5. 最长回文子串 - 动态规划、中心扩散、Manacher 算法^]
+* https://leetcode.cn/problems/maximum-length-of-repeated-subarray/solutions/28583/wu-li-jie-fa-by-stg-2/[718. 最长重复子数组 - 滑动窗口解法^]
 
 

logbook/202406.adoc

@@ -919,6 +919,10 @@
 |{doc_base_url}/0718-maximum-length-of-repeated-subarray.adoc[题解]
 |❌ 动态规划。滑动窗口的解法也很妙！
 
+|{counter:codes}
+|{leetcode_base_url}/longest-palindromic-substring/[5. Longest Palindromic Substring^]
+|{doc_base_url}/0005-longest-palindromic-substring.adoc[题解]
+|✅ 滑动窗口解法
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_0005_LongestPalindromicSubstring_3.java

@@ -0,0 +1,92 @@
+package com.diguage.algo.leetcode;
+
+import java.util.Objects;
+
+public class _0005_LongestPalindromicSubstring_3 {
+  // tag::answer[]
+
+  /**
+   * 相仿 718 题的思路。
+   *
+   * TODO 解法还有少许缺陷：142个用例，通过了140个。
+   *      abcdbbfcba 找不到合适的回文字符串。
+   *
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-10-08 14:22:42
+   */
+  public String longestPalindrome(String s) {
+    if (Objects.isNull(s) || s.length() == 1) {
+      return s;
+    }
+    return findMax(s, new StringBuilder(s).reverse().toString());
+  }
+
+  private String findMax(String a, String b) {
+    int length = b.length();
+    String result = "";
+
+    // 处理 a 的头部 与 b 的尾部 的字符串
+    for (int len = 1; len <= length; len++) {
+      String temp = maxLength(a, 0, b, length - len, len);
+      if (temp.length() > result.length() && isPalindromic(temp)) {
+        result = temp;
+      }
+    }
+    // 由于两个字符串长度相同，则不需要处理中间对其部分的处理
+    // 这里的代码假设，a 短 b 长
+    // for (int i = b.length() - a.length(); i >= 0; i--) {
+    //   String temp = maxLength(a, 0, b, i, a.length());
+    //  }
+
+    // 处理 a 的尾部 与 b 的头部
+    for (int i = 1; i < length; i++) {
+      String temp = maxLength(a, i, b, 0, length - i);
+      if (temp.length() > result.length() && isPalindromic(temp)) {
+        result = temp;
+      }
+    }
+
+    return result.isEmpty() ? null : result;
+  }
+
+  private String maxLength(String a, int as, String b, int bs, int len) {
+    int count = 0;
+    StringBuilder temp = new StringBuilder();
+    String result = "";
+    for (int i = 0; i < len; i++) {
+      if (a.charAt(as + i) == b.charAt(bs + i)) {
+        count++;
+        temp.append(a.charAt(as + i));
+        if (temp.length() > result.length()) {
+          result = temp.toString();
+        }
+      } else if (count > 0) {
+        count = 0;
+        temp = new StringBuilder();
+      }
+    }
+    int sn = a.length();
+    System.out.println(String.format("%3d %3d %3d  %" + (as + len != a.length() ? "-" : "")
+        + sn + "s  %" + (as + len == a.length() ? "-" : "") + sn + "s" + "  %" + sn + "s",
+      as, bs, len, a.substring(as, as + len), b.substring(bs, bs + len), result));
+    return result;
+  }
+
+  private boolean isPalindromic(String s) {
+    int l = 0, r = s.length() - 1;
+    while (l < r) {
+      if (s.charAt(l) != s.charAt(r)) {
+        return false;
+      }
+      l++;
+      r--;
+    }
+    return true;
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0005_LongestPalindromicSubstring_3()
+      .longestPalindrome("abcdbbfcba");
+  }
+}