Leetcode 647. Palindromic Substrings 回文子串 解题报告

这道题，就是找出一个字符串中所有可能出现的回文子串的个数。
做法嘛，就是一个个位置的统计，使用中心向外拓展的方法：
1、每个字符自己构成回文，+1
2、中心拓展，假设当前位置i为回文的中心，那么设置left=i-1 right=i+1，比较left与right位置是否相同，相同就+1，然后各自移动一步，重复直到退出
3、中心拓展当前的回文长度是偶数的，那么就设置left=i，right=i+1，其他同2一样

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:
Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".
Example 2:
Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
Note:
The input string length won't exceed 1000.

这道题我用python写的，其实用java会更简洁，因为可以（++ –）

class Solution(object):
    def countSubstrings(self, s):
        """
        :type s: str
        :rtype: int
        从中心拓展出去
        """
        res = 0
        n  = len(s)
        for i in range(0, n):
            # itself
            res += 1
            # odds
            left = i -1
            right = i + 1
            while left >= 0 and right < n and s[left] == s[right]:
                left -= 1
                right += 1
                res += 1
            # evens
            left = i
            right = i + 1
            while left >= 0 and right < n and s[left] == s[right]:
                left -= 1
                right += 1
                res += 1
        return res