Leetcode 318. Maximum Product of Word Lengths

文章作者：Tyan
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1. Description

Maximum Product of Word Lengths

2. Solution

解析： Version 1每次迭代都需要进行两次set转换，Version 2预先进行了set转换，效率有提升。由于Version 2中两个字符串的与运算非常耗时，因此Version 3先将字符串转换为代表字符串的数字，然后进行与运算，这样两个字符串的与运算就变成了两个数字按位进行与运算，大大减少了运行时间。Version 4将代表数字一样的字符串进行了合并，并保留了最大的字符串长度，这样减少了与运算的迭代次数，并且不影响最终结果。

• Version 1

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        words.sort(key=len, reverse=True)
        maximum = 0
        n = len(words)
        for i in range(n):
            for j in range(i+1, n):
                x = len(words[i])
                y = len(words[j])
                if maximum >= x * y:
                    return maximum
                intersection = set(words[i]) & set(words[j])
                if len(intersection) == 0:
                    maximum = max(maximum, x * y)
                    break
        return maximum

• Version 2

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        chars = [set(word) for word in words]
        maximum = 0
        n = len(words)
        for i in range(n):
            for j in range(i+1, n):
                x = len(words[i])
                y = len(words[j])
                intersection = chars[i] & chars[j]
                if len(intersection) == 0:
                    maximum = max(maximum, x * y)
        return maximum

• Version 3

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        lengths = [len(word) for word in words]
        flags = []
        for word in words:
            flag = 0
            for ch in word:
                flag = flag | (1 << (ord(ch) - 97))
            flags.append(flag)
        maximum = 0
        n = len(words)
        for i in range(n):
            for j in range(i+1, n):
                if flags[i] & flags[j] == 0:
                    maximum = max(maximum, lengths[i] * lengths[j])
        return maximum

• Version 4

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        maximum = 0
        mapping = {}
        for word in words:
            flag = 0
            for ch in word:
                flag = flag | (1 << (ord(ch) - 97))
            mapping[flag] = max(mapping.get(flag, 0), len(word))

        for key1, value1 in mapping.items():
            for key2, value2 in mapping.items():
                if key1 & key2 == 0:
                    maximum = max(maximum, value1 * value2)
        return maximum

Reference

1. https://leetcode.com/problems/maximum-product-of-word-lengths/