318 Maximum Product of Word Lengths解法解析
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题目为:
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.
主要有两个思路:
1.最直接的思路,利用两层遍历后找到每个单词,接着对每个单词进行比较是否存在相同字母。时间复杂度为O(n^4)。
代码如下:
public class Solution {
public int maxProduct(String[] words) {
//耗时的做法
int max=0;
for(int i=0;i<words.length;i++){
for(int j=i;j<words.length;j++){
String cur=words[i];
String mani=words[j];
HashMap<Character,Integer> curMap= new HashMap();
for(int k=0;k<cur.length();k++){
if(curMap.containsKey(cur.charAt(k)))
curMap.put(cur.charAt(k), curMap.get(cur.charAt(k))+1);
else curMap.put(cur.charAt(k), 1);
}
int flag=0;
for(int l=0;l<mani.length();l++){
if(!curMap.containsKey(mani.charAt(l)))
flag++;
}
if((flag==mani.length())&&max<=cur.length()*mani.length())
max=cur.length()*mani.length();
}
}
return max;
}
}
2.利用位操作,利用两层遍历,分两个步骤:
1)定义一个字母位数组,长度为单词个数,对26位的字母位分别赋值,字母存在则为1,字母不存在则为0。
2)两层遍历字符串数组,相与不同单词的字母位数组,结果为0则表示不存在相同的字母,结果为1则存在。
public class Solution {
public int maxProduct(String[] words) {
//利用按照位与的操作
int maniBit[]=new int[words.length];
for(int i=0;i<words.length;i++){
maniBit[i]=0;
String cur=words[i];
for(int j=0;j<cur.length();j++){
//相当于每一个数组里面保存了一个26位的一个整数,
//其中每一位代表的是每一个单词,出现则为1不出现则为0
maniBit[i]=maniBit[i]|(1<<cur.charAt(j)-'a');
}
}
int max=0;
for(int i=0;i<words.length;i++){
for(int j=i+1;j<words.length;j++){
if((maniBit[i]&maniBit[j])==0&&max<=words[i].length()*words[j].length())
max=words[i].length()*words[j].length();
}
}
return max;
}
}
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