318. 最大单词长度乘积
给定一个字符串数组 words，找到 length(word[i]) * length(word[j]) 的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。
示例 1:
输入: ["abcw","baz","foo","bar","xtfn","abcdef"]
输出: 16
解释: 这两个单词为 "abcw", "xtfn"。
示例 2:
输入: ["a","ab","abc","d","cd","bcd","abcd"]
输出: 4
解释: 这两个单词为 "ab", "cd"。
示例 3:
输入: ["a","aa","aaa","aaaa"]
输出: 0
解释: 不存在这样的两个单词。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-product-of-word-lengths
Link：https://leetcode.com/problems/maximum-product-of-word-lengths/
哈希
O(N^2)
class Solution:
def maxProduct(self, words: List[str]) -> int:
res = 0
for i in range(len(words)):
for j in range(i + 1, len(words)):
if len(set(words[i]) & set(words[j])) == 0:
res = max(res, len(words[i]) * len(words[j]))
return res
加上预处理, 速度会快一些
class Solution:
def maxProduct(self, words: List[str]) -> int:
# 预处理
word_sets = [set(word) for word in words]
res = 0
for i in range(len(words)):
for j in range(i + 1, len(words)):
if len(word_sets[i] & word_sets[j]) == 0:
res = max(res, len(words[i]) * len(words[j]))
return res
比特位
O(N^2)
因为字母只有26个，所以可以使用比特位分别代表a-z字母。
判断有没有相同字符, 按位与==0
from functools import lru_cache
class Solution:
def maxProduct(self, words: List[str]) -> int:
res = 0
for i in range(len(words)):
for j in range(i + 1, len(words)):
if (self.wordToBit(words[i]) & self.wordToBit(words[j])) == 0:
res = max(res, len(words[i]) * len(words[j]))
return res
@lru_cache(maxsize=None)
def wordToBit(self, word: str) -> int:
num = 0
for l in word:
bits = ord(l) - ord('a')
num |= 1 << bits
return num
加上预处理, 速度会快一些
class Solution:
def maxProduct(self, words: List[str]) -> int:
mask = [0 for _ in range(len(words))]
for i in range(len(words)):
for c in words[i]:
mask[i] |= 1 << (ord(c) - ord('a'))
res = 0
for i in range(len(words)):
for j in range(i + 1, len(words)):
if mask[i] & mask[j] == 0:
res = max(res, len(words[i]) * len(words[j]))
return res
–End–