[LeetCode][JavaScript]Maximum Product of Word Lengths

Maximum Product of Word Lengths

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.

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

 

 

 

 

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最直接的想法，两两比较，算出最大值。

其中比较的操作要做n^2次，只能用效率高的位运算。

先遍历一遍words，把单词转成二进制的数。

a对应2的0次，b对应2的1次，以此类推。

z	y	...	d	c	b	a
2^25	2^24	...	4	2	1	0

比较两个单词是不是包含重复字母的时候，只需要用&运算看结果是否为0。

/**
 * @param {string[]} words
 * @return {number}
 */
var maxProduct = function(words) {
    var i, j, max = 0;
    var bitDict = [];
    for(i = 0; i < words.length; i++)
        bitDict[i] = word2Num(i);
    for(i = 0; i < words.length; i++)
        for(j = 0; j < words.length; j++)
            if(i !== j && !hasSharedLetter(i, j))
                max = Math.max(max, words[i].length * words[j].length);
    return max;

    function word2Num(index){
        var res = 0, charCodeA = 'a'.charCodeAt();
        for(var i = 0 ; i < words[index].length; i++)
            res |= Math.pow(2, words[index][i].charCodeAt() - charCodeA);
        return res;
    }
    function hasSharedLetter(index1, index2){
        if((bitDict[index1] & bitDict[index2]) === 0) return false;
        return true;
    }
};