647. Palindromic Substrings 求回文,感觉有点难不知道为什么放在DP,不是用DP解
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".
class Solution(object):
def countSubstrings(self, s):
"""
:type s: str
:rtype: int
"""
n = len(s)
res = 0
for i in xrange(len(s)):
# add itself
res += 1
# even
left = i
right = i + 1
while left >= 0 and right <= n-1 and s[left] == s[right]:
res += 1
left -= 1
right += 1
# odd
left = i-1
right = i+1
while left >= 0 and right <= n-1 and s[left] == s[right]:
res += 1
left -= 1
right += 1
return res
5. Longest Palindromic Substring
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example 1:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd" Output: "bb"
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
if len(s) < 2:
return s
self.n = len(s)
self.left = len(s) - 1
self.long = 0
for i in xrange(self.n):
print i
self.helper(i, i + 1, s) # even length
self.helper(i - 1, i + 1, s) # old length
return s[self.left:self.left + self.long]
def helper(self, left, right, s):
while left >= 0 and right < self.n and s[left] == s[right]:
left -= 1
right += 1
left += 1
right -= 1
if right - left + 1 > self.long:
self.long = right - left +1
self.left = left