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描述
Given a string array words, return the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. If no such two words exist, return 0.
Example 1:
Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".
Example 2:
Input: words = ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".
Example 3:
Input: words = ["a","aa","aaa","aaaa"]
Output: 0
Explanation: No such pair of words.
Note:
2 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i] consists only of lowercase English letters.
解析
根据题意，给定一个字符串数组 words，返回 length(word[i]) * length(word[j]) 的最大值，其中两个单词不能有相同的字母。 如果不存在这样的两个词，则返回 0 。
如果按照常规的思路，我们两两进行比对，而假如两个字符串的长度为 L1 和 L2 ，要达到两个字符串中没有相同的字母，那么我们的时间复杂度为 O(N^2 * (L1 * L2)) ，这肯定是超时的，我们要保证时间复杂度能够在 O(N^2) 之内，我们肯定要降低两个字符串的比对性能 O(L1 * L2) 为 O(1) ，根据最近“每日一题”都在考位运算我们就能猜到，这道题也是在考察位运算，我们使用位运算结果来表示字符串，一共 26 个字母对应 26 个二进制位，我们只需要将每个单词表示成二进制即可，这样我们在比对两个字符串所代表的二进制是只要判断 mask[i] & mask[j] 等于 0 即可表示两个字符串没有相同的字符存在，时间复杂度可以降到 O(1) 。
整体的时间复杂度为 O(N^2) ，空间复杂度为 O(N) 。
解答
class Solution(object):
def maxProduct(self, words):
"""
:type words: List[str]
:rtype: int
"""
N = len(words)
result = 0
mask = [0] * N
for i,word in enumerate(words):
for j in range(len(word)):
mask[i] |= 1 << (ord(word[j]) - ord('a'))
for i in range(N):
for j in range(i + 1, N):
a = words[i]
b = words[j]
if not mask[i] & mask[j]:
result = max(result, len(a) * len(b))
return result
运行结果
Runtime: 499 ms, faster than 67.32% of Python online submissions for Maximum Product of Word Lengths.
Memory Usage: 14.2 MB, less than 69.91% of Python online submissions for Maximum Product of Word Lengths.
原题链接
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