leetcode 318. 最大单词长度乘积 python

题目描述：

 题解一：超时未通过

基本思路：利用set自动去重性质

1.在两个for循环中依次比较words中的两个单词word[i]和word[j]。

2.分别用两个小set保存两个单词中出现的字母，然后用一个大set保存在这两个单词中出现的所有字母，然后判断两个小set长度之和是否等于大set的长度，如果相等，说明没有重复字母，计算长度乘积。

class Solution(object):
    def maxProduct(self, words):
        word_num = len(words)
        maxmul = 0
        for i in range(word_num):
            for j in range(i+1,word_num):
                word1 = words[i]
                word2 = words[j]
                word_set1 = set()
                word_set2 = set()
                word_set = set()
                for m in range(len(word1)):
                    word_set1.add(word1[m])
                    word_set.add(word1[m])
                for n in range(len(word2)):
                    word_set2.add(word2[n])
                    word_set.add(word2[n])
                if len(word_set)==len(word_set1)+len(word_set2):
                    maxmul = max(maxmul,len(word1)*len(word2))
        return maxmul

题解二：题解一改进版 通过

1.为每个单词words[i]用一个set保存出现的字母。

2.对两个单词的set求交集，如果交集为空，则说明没有重复字母。

class Solution(object):
    def maxProduct(self, words):
        result =[]
        maxmul = 0
        word_num = len(words)
        for word in words:
            word_set = set()
            for i in range(len(word)):
                word_set.add(word[i])
            result.append(word_set)
        for i in range(word_num):
            for j in range(i+1,word_num):
                if len(result[i]&result[j])==0:
                    maxmul = max(maxmul,len(words[i])*len(words[j]))
        return maxmul

 题解三：位运算 参考

Leetcode - 318. 最大单词长度乘积 / python题解 / 判断两字符串是否存在相同元素 - 但是我拒绝 - 博客园

用一个26位的二进制数字记录每个word[i]中字母的出现情况。
对两个单词对应的二进制数字进行与运算，如果结果为0，说明没有重复。
class Solution(object):
    def maxProduct(self, words):
        result = [0] * len(words)
        maxmul = 0
        for i in range(len(words)):
            for j in words[i]:
                result[i] |= 1<<(ord(j)-97)
        for i in range(len(words)-1):
            for j in range(i+1,len(words)):
                if not (result[i]&result[j]):
                    maxmul = max(maxmul,len(words[i])*len(words[j]))
        return maxmul