Maximum Product of Word Lengths -- LeetCode Given a string arraywords, find the maximum value oflength(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....
小写字母a-z是26位,一般统计是否存在我们要申请一个bool flg[26]这样的数组,但是我们在这里用int代替,int是32位可以替代flg数组,用与(&),或(1),以及向左移位(<<)就能完成。如“abcd” 的int值为 0000 0000 0000 0000 0000 0000 0000 1111,“wxyz” 的int值为 1111 0000 0000 0000 0000 0000 0000 000...
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: Given [“abcw”, “baz”...
LeetCode 318. Maximum Product of Word Lengths 10-20 LeetCode 318. Maximum Product of Word Lengths Maximum Product of Word Lengths Description Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two wo......
简介: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。 Description Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where ...
packageleetcodefuncmaxProduct318(words[]string)int{ifwords==nil||len(words)==0{return0}length,value,maxProduct:=len(words),make([]int,len(words)),0fori:=0;i<length;i++{tmp:=words[i]value[i]=0forj:=0;j<len(tmp);j++{value[i]|=1<<(tmp[j]-'a')}}fori:=0;i<length;i+...
The product of th... YuriFLAG 0 218 [Leetcode] 58. Length of Last Word 2019-12-10 14:07 − Given a string s consisting of words and spaces, return the length of the last word in the string. A word is a maximal substring consisting of non-spa... CNoodle 0 449 Leet...
单词集合的大小,1 <= words[i].length <= 15 枚举words 子集总共 2^15 种情况 对每一种情况统计使用了哪些字母 如果字母超出范围了,就不符合要求 否则按照字母表计算得分 记录最大得分 位压缩: 对于单词集合中每一个词,都可以选择,用/不用 所以就可以用位 0/1 来表示 单词集合中每个单词都表示出来...
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