318. Maximum Product of Word Lengths

题目:

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.

链接： https://leetcode.com/problems/maximum-product-of-word-lengths/

题解：

给定一个String array，求出不含相同字符的两个单词长度乘积的最大值。

我们可以把这道题目分为几个步骤。

双重循环遍历Words

先把words[i]和words[j]放入一个<String, Set<Character>>的HashMap里

比较两者是否有重复， 假如没有重复，我们可以尝试更新max， 否则continue

最后返回max

这里题目给出单词只包含小写字母，所以我们还可以进一步优化空间复杂度。 另外，预先对数组按长度进行排序的话应该可以剪枝掉不少计算。 留给二刷了。

Java:

Time Complexity - O(n ^2)， Space Complexity - O(n)

public class Solution {
public int maxProduct(String[] words) {
if (words == null || words.length < 2) {
return 0;
}
int max = 0;
Map<String, Set<Character>> map = new HashMap<>();

for (int i = 0; i < words.length; i++) {
addToMap(words[i], map);
for (int j = i + 1; j < words.length; j++) {
addToMap(words[j], map);
if (!hasSameChar(words[i], words[j], map)) {
max = Math.max(max, words[i].length() * words[j].length());
}
}
}
return max;
}

private void addToMap(String word, Map<String, Set<Character>> map) {
if (!map.containsKey(word)) {
Set<Character> set = new HashSet<>();
for (int i = 0; i < word.length(); i++) {
set.add(word.charAt(i));
}
map.put(word, set);
}
}

private boolean hasSameChar(String word1, String word2, Map<String, Set<Character>> map) {
Set<Character> set1 = map.get(word1);
Set<Character> set2 = map.get(word2);
for (Character c : set1) {
if (set2.contains(c)) {
return true;
}
}
return false;
}
}

Reference:
https://leetcode.com/discuss/74589/32ms-java-ac-solution