318. Maximum Product of Word Lengths

Given a string array words, find themaximum
value of length(word[i]) * length(word[j]) wherethe two words do not share common letters. You may
assume that each word willcontain 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：

该题目意思是给定一个string数组，找出最大的乘积(length(word[i])*length(word[j]))，同时word[i]与word[j]没有交集。

为了判断两个字符串是否有交集，首先将数组中的字符串用一个整数表示，整数中每一位对应一个小写字母。…..xxxxxxxxxxx对应……….fedcba。两个字符串是否有交集则将对应的整数进行与运算，若结果不为0，说明有交集。

由于在寻找最大乘积时，会频繁用到字符串的长度，因此可提前使用数组保存字符串长度（该过程在一定程度上会减少运行时间）

使用指针指向固定大小的数组相比于使用vector划分固定大小的数组，运行时间较少。

#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
class Solution {
public:
int maxProduct(vector<string>& words) {
int n=words.size();
if(n<2)return 0;
// 'vector<int>res' costs 20ms more than using 'int*res=new int
'
//vector<int>res=vector<int>(n,0);
int*res=new int
;
// using array len to store the length of words,it can improve the time.
int*len=new int
;
for(int i=0;i<n;i++){
//string word=words[i];
//transform(word.begin(),word.end(),word.begin(),::tolower);
res[i]=0;
len[i]=words[i].size();
int l=words[i].size();
for(int j=0;j<l;j++){
res[i]|=1<<(words[i][j]-'a');
}
//res[i]=bit;
//cout<<bit<<endl;
}
int max=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++)
if(!(res[i]&res[j])){
int tmp=len[i]*len[j];
if(max<tmp)max=tmp;
}
}
return max;
}
};
//int main(){
//  vector<string>v;
//  int n;
//  cin>>n;
//  for(int i=0;i<n;i++){
//      string tmp;
//      cin>>tmp;
//      v.push_back(tmp);
//  }
//  Solution test=Solution();
//  cout<<test.maxProduct(v)<<endl;
//  system("pause");
//  return 0;
//}