golang 解法, 执行用时: 60 ms, 内存消耗: 7.1 MB, 提交时间: 2023-06-07 10:50:03

func wordSubsets(words1 []string, words2 []string) []string {
    record := make([]int,26)
    ans := make([]string,0)
    for _,i :=range words2{
        tmp := make([]int,26)
        for _,j:=range i{
            tmp[j-97] += 1
        }
        for i:=0;i<26;i++{
            record[i] = max(record[i],tmp[i])
        }
    }
    for _,i := range words1{
        if isValid(i,record){
            ans = append(ans,i)
        }
    }
    return ans
}
func max(a,b int)int{
    if a>b{
        return a
    }
    return b
}
func isValid(arr string,b []int) bool{
    a := make([]int,26)
    for _,j:=range arr{
        a[j-97] += 1
    }
    for i:=0;i<26;i++{
        if b[i]>a[i]{
            return false
        }
    }
    return true
}

python3 解法, 执行用时: 800 ms, 内存消耗: 19.5 MB, 提交时间: 2023-06-07 10:48:59

'''
1.合B所有单词之力，统计26个字母的最高频次。
2.A中的a，要想符合，就必须26个字母每个字母的freq都超过B中的
'''
class Solution:
    def wordSubsets(self, A: List[str], B: List[str]) -> List[str]:
        B_char_freq = [0 for _ in range(26)]
        for b in B:
            tmp = self.func_count(b)
            for i in range(26):
                B_char_freq[i] = max(B_char_freq[i], tmp[i])
        
        res = []
        for a in A:
            tmp = self.func_count(a)
            #if all (x >= y for x,y in zip(tmp, B_char_freq)):
            if all (tmp[i] >= B_char_freq[i] for i in range(26)):
                res.append(a)
        return res
    

    def func_count(self, s: str) -> List[int]:
        char_freq = [0 for _ in range(26)]
        for c in s:
            char_freq[ord(c) - ord('a')] += 1
        return char_freq

python3 解法, 执行用时: 744 ms, 内存消耗: 19.5 MB, 提交时间: 2023-06-07 10:47:14

class Solution:
    def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
        def count(word: str) -> str:
            ans = [0] * 26
            for letter in word:
                ans[ord(letter) - ord('a')] += 1
            return ans

        bmax = [0] * 26
        for b in words2:
            for i, c in enumerate(count(b)):
                bmax[i] = max(bmax[i], c)

        ans = []
        for a in words1:
            if all(x >= y for x, y in zip(count(a), bmax)):
                ans.append(a)
        return ans

java 解法, 执行用时: 15 ms, 内存消耗: 51.6 MB, 提交时间: 2023-06-07 10:45:12

// 将words2合并成一个单词
class Solution {
    public List<String> wordSubsets(String[] A, String[] B) {
        int[] bmax = count("");
        for (String b: B) {
            int[] bCount = count(b);
            for (int i = 0; i < 26; ++i)
                bmax[i] = Math.max(bmax[i], bCount[i]);
        }

        List<String> ans = new ArrayList();
        search: for (String a: A) {
            int[] aCount = count(a);
            for (int i = 0; i < 26; ++i)
                if (aCount[i] < bmax[i])
                    continue search;
            ans.add(a);
        }

        return ans;
    }

    public int[] count(String S) {
        int[] ans = new int[26];
        for (char c: S.toCharArray())
            ans[c - 'a']++;
        return ans;
    }
}