Leetcode 318.最大单词长度乘积

【Leetcode 318.最大单词长度乘积】的更多相关文章

Java实现 LeetCode 318 最大单词长度乘积

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 &…

Leetcode 318.最大单词长度乘积

最大单词长度乘积 . 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…

[Leetcode] 第318题最大单词长度乘积

一.题目描述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 "ab…

318 Maximum Product of Word Lengths 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,并且两个单词不含公共的字母.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"]返回 16两个单词可以为 "abcw", "…

[Swift]LeetCode318. 最大单词长度乘积 | Maximum Product of Word Lengths

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…

[leetcode]318. Maximum Product of Word Lengths单词长度最大乘积

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…

[LeetCode] Maximum Product of Word Lengths 单词长度的最大积

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…