2085. Count Common Words With One Occurrence-统计出现过一次的公共字符串

给你两个字符串数组words1和words2，请你返回在两个字符串数组中都恰好出现一次的字符串的数目。

示例 1：

输入：words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"]
输出：2
解释：
- "leetcode" 在两个数组中都恰好出现一次，计入答案。
- "amazing" 在两个数组中都恰好出现一次，计入答案。
- "is" 在两个数组中都出现过，但在 words1 中出现了 2 次，不计入答案。
- "as" 在 words1 中出现了一次，但是在 words2 中没有出现过，不计入答案。
所以，有 2 个字符串在两个数组中都恰好出现了一次。

示例 2：

输入：words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"]
输出：0
解释：没有字符串在两个数组中都恰好出现一次。

示例 3：

输入：words1 = ["a","ab"], words2 = ["a","a","a","ab"]
输出：1
解释：唯一在两个数组中都出现一次的字符串是 "ab" 。

提示：

• 1 <= words1.length, words2.length <= 1000
• 1 <= words1[i].length, words2[j].length <= 30
• words1[i]和words2[j]都只包含小写英文字母。

Python:

class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        adic, bdic = dict(), dict()
        total = 0
        for item in words1:
            if item not in adic.keys():
                adic[item] = 1
            else:
                adic[item] += 1
        for item in words2:
            if item not in bdic.keys():
                bdic[item] = 1
            else:
                bdic[item] += 1
        for key in adic.keys():
            if adic[key] == 1 and key in bdic.keys() and bdic[key] == 1:
                total += 1
        return total

Java:

class Solution {
    public int countWords(String[] words1, String[] words2) {
        int total = 0;
        Map<String, Integer> a = new HashMap<>();
        Map<String, Integer> b = new HashMap<>();
        for(String item: words1)
        {
            if(!a.containsKey(item))
            {
                a.put(item, 1);
            }
            else
            {
                a.compute(item, (k, v)-> v+=1);
            }
        }
        for(String item: words2)
        {
            if(!b.containsKey(item))
            {
                b.put(item, 1);
            }
            else
            {
                b.compute(item, (k, v)-> v+=1);
            }
        }
        for(String key: a.keySet())
        {
            if(b.containsKey(key) && a.get(key)== 1 && b.get(key) == 1)
                total += 1;
        }
        return total;
    }
}