码迷,mamicode.com
首页 > 其他好文 > 详细

647. Palindromic Substrings 回文子串的数量

时间:2017-08-13 22:13:45      阅读:594      评论:0      收藏:0      [点我收藏+]

标签:order   string   ons   不同的   ==   family   start   substr   height   

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:

  1. The input string length won‘t exceed 1000.
求回文子串的数量,每一个子串拥有不同的star index 和不同的end index
  1. class Solution(object):
  2. def countSubstrings(self, s):
  3. """
  4. :type s: str
  5. :rtype: int
  6. """
  7. num = 0
  8. for i in range(0, len(s) + 1):
  9. for j in range(i + 1, len(s) + 1):
  10. sub = s[i:j]
  11. if(sub == sub[::-1]):
  12. num = num + 1
  13. return num






647. Palindromic Substrings 回文子串的数量

标签:order   string   ons   不同的   ==   family   start   substr   height   

原文地址:http://www.cnblogs.com/xiejunzhao/p/7354950.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!