318. Maximum Product of Word Lengths | 字符串长度相乘最大值

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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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思路：创建一个int数组，长度为string数组的长度，其中每个数对应该字符串中对应位置的字符串，值是26位，最低位为a对应，最高位对应z，当该字母存在时，则对应位位1，体现在代码中则为value[i] |= 1 << (tmp.charAt(j) - 'a');随后再遍历字符串数组，当两个字符串对应的value值相与为0,(若存在相同的字符，则相与会大于0)且长度相乘大于当前最大值，遍历完返回最大值即可。

public class Solution {    public int maxProduct(String[] words) {if (words.length == 0 || words.length == 1) {return 0;}int len = words.length;int[] value = new int[len];for (int i = 0; i < len; i++) {String tmp = words[i];value[i] = 0;for (int j = 0; j < tmp.length(); j++) {value[i] |= 1 << (tmp.charAt(j) - 'a');}}int maxProduct = 0;for (int i = 0; i < len - 1; i++) {for (int j = i + 1; j < len; j++) {if (((value[i] & value[j]) == 0) && words[i].length() * words[j].length() > maxProduct) {maxProduct = words[i].length() * words[j].length();}}}return maxProduct;}}