add 647. Palindromic Substrings.md 🍺

README.md

@@ -325,6 +325,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 621 |[Task Scheduler](https://leetcode.com/problems/task-scheduler/)|[C++](solutions/621.%20Task%20Scheduler.md)|Medium|  |
 | 628 |[Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/628.%20Maximum%20Product%20of%20Three%20Numbers.md)|Easy|  |
 | 643 |[Maximum Average Subarray I](https://leetcode.com/problems/maximum-average-subarray-i)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/643.%20Maximum%20Average%20Subarray%20I.md)|Easy|  |
+| 647 |[Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)|[C++](solutions/647.%20Palindromic%20Substrings.md)|Medium|  |
 | 661 |[Image Smoother](https://leetcode.com/problems/image-smoother)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/661.%20Image%20Smoother.md)|Easy|  |
 | 665 |[Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/665.%20Non-decreasing%20Array.md)|Easy|  |
 | 714 |[Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee.md)|Medium|  |

solutions/647. Palindromic Substrings.md

@@ -0,0 +1,69 @@
+# [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)
+
+# 思路
+计算有多少个回文子串。
+
+## 思路一、扩散法
+
+根据回文串的定义，回文串是对称的。所以我们可以以字符串中的每个字符作为回文串中心位置，然后向两边扩散，每当成功匹配两个左右两个字符，就说明找到了一个回文串，res自增1。注意回文字符串有奇数和偶数两种形式，处理方式略有不同。
+
+空间复杂度O(1)， 时间复杂度O(n^2)
+
+## 思路二、动态规划
+
+还可以用动归来做：
+```
+dp[i][j] (i <= j) 定义成子字符串s[i,...,j]是否是回文串
+```
+所以初始状态就是`dp[i][i] = true`，状态转移方程为：
+```
+if s[i] == s[j] && (i+1 == j || dp[i+1][j-1]) :
+    dp[i][j] = true
+```
+
+空间复杂度O(n^2)， 时间复杂度O(n^2)，亲测比思路一慢不少
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    int countSubstrings(string s) {
+        int n = s.size(), res = 0;
+        
+        for(int i = 0; i < n; i++){
+            // 奇
+            for(int j = 0; j <= min(i, n - 1 - i); j++){
+                if(s[i-j] == s[i+j]) res++;
+                else break;
+            }
+            // 偶
+            for(int j = 1; j <= min(i+1, n - 1 - i); j++){
+                if(s[i-j+1] == s[i+j]) res++;
+                else break;
+            }
+        }
+        return res;
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+public:
+    int countSubstrings(string s) {
+        int n = s.size(), res = 0;
+        
+        vector<vector<bool>>dp(n, vector<bool>(n, false));
+        for(int i = n - 1; i >= 0; i--)
+            for(int j = n - 1; j >= i; j--){
+                if(s[i] == s[j] && (i == j || i+1 == j || dp[i+1][j-1])){
+                    dp[i][j] = true;
+                    res++;
+                }
+            }
+        return res;
+    }
+};
+```