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", "…

传送门 318. Maximum Product of Word Lengths My Submissions QuestionEditorial Solution Total Accepted: 19855 Total Submissions: 50022 Difficulty: Medium Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two…

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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 set 位运算 日期 题目地址:https://leetcode.com/problems/maximum-product-of-word-lengths/description/ 题目描述 Given a string array words, find the maximum value of length(word[i]) * length(w…

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…

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…

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…

题目描述: 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. E…

https://leetcode.com/problems/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 lo…

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…

题目大意:给出一些字符串,找出两个不同的字符串之间长度之积的最大值,但要求这两个字符串之间不能拥有相同的字符.(字符只考虑小写字母). 题目分析:字符最多只有26个,因此每个字符串可以用一个二进制数来表示它含有哪些字符. 代码如下: class Solution { public: int maxProduct(vector<string>& words) { int maxn=0; int *Bit=new int[words.size()]; fill(Bit,Bit+words.…

先介绍下本题的题意: 在一个字符串组成的数组words中,找出max{Length(words[i]) * Length(words[j]) },其中words[i]和words[j]中没有相同的字母,在这里字符串由小写字母a-z组成的. 对于这道题目我们统计下words[i]的小写字母a-z是否存在,然后枚举words[i]和words[j],找出max{Length(words[i]) * Length(words[j]) }. 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…

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…

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…

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…

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…

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…

Description: 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, retu…

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 &…

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…

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…

题目链接:https://leetcode.com/problems/maximum-product-subarray/description/ 题目大意:给出一串数组,找出连续子数组中乘积最大的子数组的乘积. 法一:暴力.竟然能过,数据也太水了.两个for循环,遍历每一个可能的连续子数组,找出最大值.代码如下(耗时161ms): public int maxProduct(int[] nums) { int res = Integer.MIN_VALUE, len = nums.length;…

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Given a sequence of words, check whether it forms a valid word square. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤k < max(numRows, numColumns). Note: The number of words given is at le…

Find the contiguous subarray within an array (containing at least one number) which has the largest product. For example, given the array [2,3,-2,4],the contiguous subarray [2,3] has the largest product = 6. 这个求最大子数组乘积问题是由最大子数组之和问题演变而来,但是却比求最大子数组之和要复…

题目链接:Maximum Product Subarray solutions同步在github 题目很简单,给一个数组,求一个连续的子数组,使得数组元素之积最大.这是求连续最大子序列和的加强版,我们可以先看看求连续最大子序列和的题目maximum-subarray,这题不难,我们举个例子. 假设数组[1, 2, -4, 5, -1, 10],前两个相加后得到3,更新最大值(为3),然后再加上-4后,和变成-1了,这时我们发现如果-1去加上5,不如舍弃前面相加的sum,5单独重新开始继续往后相加…

注意long long  long long  longlong !!!!!!   还有 printf的时候 明明longlong型的答案 用了%d  WA了也看不出,这个细节要注意!!! #include <cstdio> ]; int main() { ;long long ans,sum; while(~scanf("%d",&n)) { ;i<n;i++) scanf("%d",&a[i]); ans=; ;i<n;i…

Maximum Product Subarray Title: Find the contiguous subarray within an array (containing at least one number) which has the largest product. For example, given the array [2,3,-2,4],the contiguous subarray [2,3] has the largest product = 6. 对于Product…

Problem D - Maximum Product Time Limit: 1 second Given a sequence of integers S = {S1, S2, ..., Sn}, you should determine what is the value of the maximum positive product involving consecutive terms ofS. If you cannot find a positive sequence, you s…

最大乘积 Maximum Product 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=84562#problem/B 解题思路: 题目意思是输入n个元素组成的序列S,找出一个乘积最大的连续子序列.若这个数不是正数,则输出0(表示无解).分析 ,连续子序列有两个要素:起点和终点,因此只需要枚举起点和终点即可.分析最大可能的乘积不会超过10的18次方,所以用 long long 来存储即可. 程序代码: #include <c…