palindrome-partitioning-ii

最新推荐文章于 2019-01-09 21:51:29 发布

原创最新推荐文章于 2019-01-09 21:51:29 发布 · 213 阅读

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#算法 python #string

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本文介绍了一个算法问题：如何计算将字符串分割成多个回文子串所需的最少切割次数。通过动态规划的方法，文章给出了一个高效的解决方案，并附带了实现代码。

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s ="aab",

Return1since the palindrome partitioning["aa","b"]could be produced using 1 cut.

def minCut(S):
   dp = [len(S) for _ in S]
   print dp
   index = 0
   while index < len(S):
      if isPalindrome(S[:index + 1]):
         dp[index] = 0
      j = 1
      while j <= index:
         if isPalindrome(S[j:index + 1]):
            dp[index] = min(dp[index], dp[j - 1] + 1)
         else:
            dp[index] = min(dp[index], dp[j - 1] + index - j + 1)
         j += 1
      index += 1
   return dp[len(S) - 1]

def isPalindrome(s):
   i = 0
   j = len(s) - 1
   while i < j:
      if s[i] == s[j]:
         i += 1
         j -= 1
      else:
         return False
   return True

print minCut("abaccddgerg")