【leetcode】647. 回文子串（python，动态规划法；中心拓展法；暴力破解）

class Solution:
    def countSubstrings(self, s: str) -> int:
        cnt = 0
        n = len(s)
        dp = [[False for i in range(n)] for j in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i, n):
                if s[i] == s[j] and (j - i <= 1 or dp[i + 1][j - 1]):
                    cnt += 1
                    dp[i][j] = True
        return cnt

class Solution:
    def countSubstrings(self, s: str) -> int:
        def speard(l, r):
            cnt = 0
            while l >= 0 and r < len(s):
                if s[l] == s[r]:
                    # print(s[l: r + 1]) # 左闭右开
                    l -= 1
                    r += 1
                    cnt += 1
                else:
                    break
            return cnt
        
        ans = 0
        for i in range(len(s)):
            ans += speard(i, i) # 中心是一个字符；
            ans += speard(i, i + 1) # 中心是2个字符；
        return ans

class Solution:
    def isPalindrome(self, s, low, high):
        return s[low: high] == s[high: low: -1]        
    def countSubstrings(self, s: str) -> int:   # 暴力解法
        cnt = 0
        for i in range(len(s)):
            for j in range(i + 1):
                if self.isPalindrome(s, j, i):
                    cnt += 1
        return cnt