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 [[]]