From e9274b0a6ba153bbafd5a82d9a37a94d0e8a7147 Mon Sep 17 00:00:00 2001
From: ZhaoJianyu-trhy <741766685@qq.com>
Date: Mon, 17 May 2021 22:39:54 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A00647=E5=9B=9E=E6=96=87?=
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---
 problems/0647.回文子串.md | 61 ++++++++++++++++++++++++++++++++++-
 1 file changed, 60 insertions(+), 1 deletion(-)

diff --git a/problems/0647.回文子串.md b/problems/0647.回文子串.md
index 45bb72be..bfc52502 100644
--- a/problems/0647.回文子串.md
+++ b/problems/0647.回文子串.md
@@ -219,9 +219,68 @@ public:
 
 ## 其他语言版本
 
-
 Java：
 
+动态规划：
+
+```java
+class Solution {
+    public int countSubstrings(String s) {
+        int len, ans = 0;
+        if (s == null || (len = s.length()) < 1) return 0;
+        //dp[i][j]：s字符串下标i到下标j的字串是否是一个回文串，即s[i, j]
+        boolean[][] dp = new boolean[len][len];
+        for (int j = 0; j < len; j++) {
+            for (int i = 0; i <= j; i++) {
+                //当两端字母一样时，才可以两端收缩进一步判断
+                if (s.charAt(i) == s.charAt(j)) {
+                    //i++，j--，即两端收缩之后i,j指针指向同一个字符或者i超过j了,必然是一个回文串
+                    if (j - i < 3) {
+                        dp[i][j] = true;
+                    } else {
+                        //否则通过收缩之后的字串判断
+                        dp[i][j] = dp[i + 1][j - 1];
+                    }
+                } else {//两端字符不一样，不是回文串
+                    dp[i][j] = false;
+                }
+            }
+        }
+        //遍历每一个字串，统计回文串个数
+        for (int i = 0; i < len; i++) {
+            for (int j = 0; j < len; j++) {
+                if (dp[i][j]) ans++;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+中心扩散法：
+
+```java
+class Solution {
+    public int countSubstrings(String s) {
+        int len, ans = 0;
+        if (s == null || (len = s.length()) < 1) return 0;
+        //总共有2 * len - 1个中心点
+        for (int i = 0; i < 2 * len - 1; i++) {
+            //通过遍历每个回文中心，向两边扩散，并判断是否回文字串
+            //有两种情况，left == right，right = left + 1，这两种回文中心是不一样的
+            int left = i / 2, right = left + i % 2;
+            while (left >= 0 && right < len && s.charAt(left) == s.charAt(right)) {
+                //如果当前是一个回文串，则记录数量
+                ans++;
+                left--;
+                right++;
+            }
+        }
+        return ans;
+    }
+}
+```
+
 
 Python：