From b934f47c83340b91b8a6291d96420978da0ec0e6 Mon Sep 17 00:00:00 2001
From: milu-tao <91822069+milu-tao@users.noreply.github.com>
Date: Fri, 24 Mar 2023 22:14:39 +0800
Subject: [PATCH] =?UTF-8?q?Update=200647.=E5=9B=9E=E6=96=87=E5=AD=90?=
 =?UTF-8?q?=E4=B8=B2.md?=
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1. result 在遍历过程中 ++ 就行，没必要另外计算
2.题目已经表明：s字符串的长度规则是：1 <= s.length <= 1000，所以前面不需要进行判断字符串长度
---
 problems/0647.回文子串.md | 33 ++++++++++++---------------------
 1 file changed, 12 insertions(+), 21 deletions(-)

diff --git a/problems/0647.回文子串.md b/problems/0647.回文子串.md
index 90e6da9f..339247f5 100644
--- a/problems/0647.回文子串.md
+++ b/problems/0647.回文子串.md
@@ -238,33 +238,24 @@ Java：
 ```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]
+        char[] chars = s.toCharArray();
+        int len = chars.length;
         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) {
+        int result = 0;
+        for (int i = len - 1; i >= 0; i--) {
+            for (int j = i; j < len; j++) {
+                if (chars[i] == chars[j]) {
+                    if (j - i <= 1) { // 情况一 和 情况二
+                        result++;
+                        dp[i][j] = true;
+                    } else if (dp[i + 1][j - 1]) { //情况三
+                        result++;
                         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;
+        return result;
     }
 }
 ```