Palindrome Partitioning 解答

Question

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

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

Solution

基本思路还是递归。每次只探究第一刀切在哪儿。这里，为了避免重复计算，我们用DP来先处理子串是否对称的问题。思路参见 Palindrom Subarrays

如果想进一步节省时间，可以参见Word Break II的解法，将每个子问题的解存起来。

 class Solution(object):
     def construction(self, s):
         length = len(s)
         self.dp = [[False for i in range(length)] for j in range(length)]
         for i in range(length):
             self.dp[i][i] = True
         for i in range(length - 1):
             if s[i] == s[i + 1]:
                 self.dp[i][i + 1] = True
         for sub_len in range(3, length + 1):
             for start in range(0, length - sub_len + 1):
                 end = start + sub_len - 1
                 if s[start] == s[end] and self.dp[start + 1][end - 1]:
                     self.dp[start][end] = True

     def partition(self, s):
         """
         :type s: str
         :rtype: List[List[str]]
         """
         self.construction(s)
         result = []
         self.helper(s, 0, [], result)
         return result

     def helper(self, s, start, cur_list, result):
         length = len(s)
         if start == length:
             result.append(list(cur_list))
             return
         for end in range(start, length):
             if self.dp[start][end]:
                 cur_list.append(s[start : end + 1])
                 self.helper(s, end + 1, cur_list, result)
                 cur_list.pop()
     

由于Python本身对字符串的强大处理，这道题的解答也可以为：

（比上一个解法花时间多）

 class Solution(object):
     def partition(self, s):
         """
         :type s: str
         :rtype: List[List[str]]
         """
         return [[s[:i]] + rest
             for i in xrange(1, len(s)+1)
             if s[:i] == s[i-1::-1]
             for rest in self.partition(s[i:])] or [[]]