318. Maximum Product of Word Lengths
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Difficulty:Medium
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", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn"
.
Example 2:
Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd"
.
Example 3:
Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.
解题思路:
对于一个字符串向量,题目描述中可见对每个字符串自身的字母排序是没有要求的,所以可以考虑使用另一种方式存贮每一个字符串。所以按照从0-26的位置存贮a-z的字母,所以开出一个二维数组,数组大小为arr[n][26],其中n表示字符串的个数,26用来存储每个字符串中字母的个数。
之后利用一个三层循环,最外层控制第一个字符串的位置,用第二层循环控制第二个字符串的位置,第三层循环26次,判断两个字符串中相应位置每个字母个数相乘的结果,最后将这26个数值相加,如果两个字符串中有字母重复,那么就会相乘结果大于等于1,所以最终结果的判断是否结果等于一并且判断那两个组合有最大的结果
class Solution {public: int maxProduct(vector<string>& words) { int n=words.size();int arr[n][26]={};for(int i=0;i<n;i++){int size=words[i].size();string s=words[i];for(int j=0;j<size;j++){arr[i][s[j]-'a']++;}}int first=0;int second=0;int max=0;for(int i=0;i<n-1;i++){for(int j=i+1;j<n;j++){bool judge=0;int sum=0;for(int k=0;k<26;k++){sum+=arr[i][k]*arr[j][k];}if(sum==0) {if(max<words[i].size()*words[j].size())max=words[i].size()*words[j].size();}}}return max; }};
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