Go Java算法最大单词长度乘积示例详解

class Solution {
    public int maxProduct(String[] words) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        int length = words.length;
        for (int i = 0; i < length; i++) {
            int mask = 0;
            String word = words[i];
            int wordLength = word.length();
            for (int j = 0; j < wordLength; j++) {
                mask |= 1 << (word.charAt(j) - 'a');
            }
            if (wordLength > map.getOrDefault(mask, 0)) {
                map.put(mask, wordLength);
            }
        }
        int maxProd = 0;
        Set<Integer> maskSet = map.keySet();
        for (int mask1 : maskSet) {
            int wordLength1 = map.get(mask1);
            for (int mask2 : maskSet) {
                if ((mask1 & mask2) == 0) {
                    int wordLength2 = map.get(mask2);
                    maxProd = Math.max(maxProd, wordLength1 * wordLength2);
                }
            }
        }
        return maxProd;
    }
}

func maxProduct(words []string) int {
    hash := func(word string) int {
        res := 0
        for _, r := range word{
            res |= 1 << (r - 'a')
        }
        return res
    }
    m, ans := map[int]int{}, 0
    for _, word := range words {
        h := hash(word)
        if m[h] < len(word) {
            for other, v := range m {
                if((other & h) == 0){
                    if tmp := v * len(word); tmp > ans {
                        ans = tmp
                    }
                }
            }
            m[h] = len(word)
        }
    }
    return ans
}