求教leetcode上Palindrome Partitioning DFS解法的复杂度# JobHunting - 待字闺中
r*u
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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"]
]
DFS就是要枚举所有可能的划分，然后检查是否是palindrome吧，中间如果发现划分不
满足palindrome了，就backtrack。这个复杂度应该怎么求？
a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab",
Return
[
["aa","b"],
["a","a","b"]
]
DFS就是要枚举所有可能的划分，然后检查是否是palindrome吧，中间如果发现划分不
满足palindrome了，就backtrack。这个复杂度应该怎么求？
