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"
最长回文字符串，标准动态规划，找到转移点“如果一个字符串是回文串，那么在它左右分别加上一个相同的字符，那么它一定还是一个回文串”；dp关键是两部分，一个是base case，一个转移方阵；base case有两种情况，单个字符和两个相邻相同字符；使用dp[i][j]表示从i位置到j位置的字符串是否是回文。
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
if len(s) <= 1:
return s
dp = [[False for i in range(len(s))] for j in range(len(s))]
res = ""
for i in range(len(s)-1, -1, -1):
for j in range(i, len(s)):
if i == j:
dp[i][j] = True
elif abs(i-j) == 1 and s[i] == s[j]:
dp[i][j] = True
elif s[i] == s[j] and dp[i+1][j-1] == True:
dp[i][j] = True
tmp = s[i:j+1]
if dp[i][j] == True and len(tmp) > len(res):
res = tmp
return res
32. Longest Valid Parentheses
Given a string containing just the characters ‘(’ and ‘)’, find the length of the longest valid (well-formed) parentheses substring.
Example 1:
Input: "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()"
Example 2:
Input: ")()())"
Output: 4
Explanation: The longest valid parentheses substring is "()()"
用一维数组来表示，不要用二维的布尔矩阵表示
) ( ) ( ( ) ) )
0 1 2 3 4 5 6 7
看当前括号的前一个括号的匹配情况，例如在7之前以6结尾的的最佳匹配是3-6，看3之前的括号和7是否匹配，不匹配则没有变化；而6之前以5结尾的最佳匹配是4-5，此时3和6匹配，则dp[i]+2。此外，如果与当前括号匹配的左括号之前的括号的dp值也应该加进来，因为由于添加了当前的括号，那些括号也被连接起来了。例如3和6匹配后，1和2也应该被加到以6结尾的最佳匹配中。
class Solution(object):
def longestValidParentheses(self, s):
"""
:type s: str
:rtype: int
"""
if len(s) <=1:
return 0
length = len(s)
dp = [0 for i in range(length)]
for i in range(1, length):
if s[i] == ")":
j = i - 1 - dp[i - 1]
if j >= 0 and s[j] == "(":
dp[i] = dp[i - 1] + 2
if j - 1 >= 0:
dp[i] += dp[j - 1]
return max(dp)