题目地址：https://leetcode.com/problems/maximum-product-of-word-lengths/description/
题目描述
Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.
Example 1:
Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".
Example 2:
Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".
Example 3:
Input: ["a","aa","aaa","aaaa"]
Output: 0
Explanation: No such pair of words.
题目大意
找出两个不包含重复字符的字符串长度乘积最大值。
解题方法
set
重点是不包含重复字符，显然可以用set统计每个字符串中出现的字符，然后利用O(n^2)的时间复杂度暴力求解，竟然过了！
Python代码如下：
class Solution:
def maxProduct(self, words):
"""
:type words: List[str]
:rtype: int
"""
word_dict = dict()
for word in words:
word_dict[word] = set(word)
max_len = 0
for i1, w1 in enumerate(words):
for i2 in range(i1+1, len(words)):
w2 = words[i2]
if not (word_dict[w1] & word_dict[w2]):
max_len = max(max_len, len(w1) * len(w2))
return max_len
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位运算
这个是个巧妙的方法。我们知道int有32位，而英文小写字符只有26个，所以，对于一个字符串，把其出现过的字符对应到int上去，那么这个int就能当做这个字符串的摘要，表示这个这个字符串中都有哪些字符。
我们把每个字符串都形成一个摘要，这样只要两个字符串的摘要想与之后的结果是0，那么说明两个字符串没有公共字符。
是不是感觉这个方法似曾相识呢？没错，这个很类似布隆过滤器open in new window啊！
python代码如下：
class Solution(object):
def maxProduct(self, words):
"""
:type words: List[str]
:rtype: int
"""
res = 0
d = collections.defaultdict(int)
N = len(words)
for i in range(N):
w = words[i]
for c in w:
d[w] |= 1 << (ord(c) - ord('a'))
for j in range(i):
if not d[words[j]] & d[words[i]]:
res = max(res, len(words[j]) * len(words[i]))
return res
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参考资料：http://www.cnblogs.com/grandyang/p/5090058.html
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