feat: add 005

Author: Blankj

README.md

@@ -69,6 +69,7 @@
 |:------------- |:------------- |:------------- |
 |2|[Add Two Numbers][002]|Linked List, Math|
 |3|[Longest Substring Without Repeating Characters][003]|Hash Table, Two Pointers, String|
+|5|[Longest Palindromic Substring][005]|String|
 |8|[String to Integer (atoi)][008]|Math, String|
 |15|[3Sum][015]|Array, Two Pointers|
 |17|[Letter Combinations of a Phone Number][017]|String, Backtracking|

note/005/README.md

@@ -0,0 +1,77 @@
+# [Longest Palindromic Substring][title]
+
+## Description
+
+Given a string **s**, find the longest palindromic substring in **s**. You may assume that the maximum length of **s** is 1000.
+
+**Example:**
+
+```
+Input: "babad"
+
+Output: "bab"
+
+Note: "aba" is also a valid answer.
+
+```
+
+**Example:**
+
+```
+Input: "cbbd"
+
+Output: "bb"
+```
+
+**Tags:** String
+
+
+## 思路0
+
+题意是寻找出字符串中最长的回文串，所谓回文串就是正序和逆序相同的字符串，也就是关于中间对称。我们先用最常规的做法，依次去求得每个字符的最长回文，要注意每个字符有奇数长度的回文串和偶数长度的回文串两种情况，相信你可以很轻易地从如下代码中找到相关代码，记录最长回文的始末位置即可，时间复杂度的话，首先要遍历一遍字符串，然后对每个字符都去求得最长回文，所以时间复杂度为O(n^2)。
+
+```java
+class Solution {
+ int st, end;
+
+ public String longestPalindrome(String s) {
+ int len = s.length();
+ if (len <= 1) return s;
+ char[] chars = s.toCharArray();
+ for (int i = 0; i < len; i++) {
+ helper(chars, i, i);
+ helper(chars, i, i + 1);
+ }
+ return s.substring(st, end + 1);
+ }
+
+ private void helper(char[] chars, int l, int r) {
+ while (l >= 0 && r < chars.length && chars[l] == chars[r]) {
+ --l;
+ ++r;
+ }
+ if (end - st < r - l - 2) {
+ st = l + 1;
+ end = r - 1;
+ }
+ }
+}
+```
+
+## 思路1
+
+
+
+```java
+
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/longest-palindromic-substring
+[ajl]: https://github.com/Blankj/awesome-java-leetcode

src/com/blankj/medium/_005/Solution.java

@@ -0,0 +1,41 @@
+package com.blankj.medium._005;
+
+/**
+ * <pre>
+ * author: Blankj
+ * blog : http://blankj.com
+ * time : 2017/11/04
+ * desc :
+ * </pre>
+ */
+public class Solution {
+ int st, end;
+
+ public String longestPalindrome(String s) {
+ int len = s.length();
+ if (len <= 1) return s;
+ char[] chars = s.toCharArray();
+ for (int i = 0; i < len; i++) {
+ helper(chars, i, i);
+ helper(chars, i, i + 1);
+ }
+ return s.substring(st, end + 1);
+ }
+
+ private void helper(char[] chars, int l, int r) {
+ while (l >= 0 && r < chars.length && chars[l] == chars[r]) {
+ --l;
+ ++r;
+ }
+ if (end - st < r - l - 2) {
+ st = l + 1;
+ end = r - 1;
+ }
+ }
+
+ public static void main(String[] args) {
+ Solution solution = new Solution();
+ System.out.println(solution.longestPalindrome("babad"));
+ System.out.println(solution.longestPalindrome("cbbd"));
+ }
+}