LeetCode2085. Count Common Words With One Occurrence

文章目录

一、题目

Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays.

Example 1:

Input: words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"]

Output: 2

Explanation:

  • "leetcode" appears exactly once in each of the two arrays. We count this string.
  • "amazing" appears exactly once in each of the two arrays. We count this string.
  • "is" appears in each of the two arrays, but there are 2 occurrences of it in words1. We do not count this string.
  • "as" appears once in words1, but does not appear in words2. We do not count this string.
    Thus, there are 2 strings that appear exactly once in each of the two arrays.
    Example 2:

Input: words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"]

Output: 0

Explanation: There are no strings that appear in each of the two arrays.

Example 3:

Input: words1 = ["a","ab"], words2 = ["a","a","a","ab"]

Output: 1

Explanation: The only string that appears exactly once in each of the two arrays is "ab".

Constraints:

1 <= words1.length, words2.length <= 1000

1 <= words1[i].length, words2[j].length <= 30

words1[i] and words2[j] consists only of lowercase English letters.

二、题解

cpp 复制代码
class Solution {
public:
    int countWords(vector<string>& words1, vector<string>& words2) {
        unordered_map<string,int> map1;
        unordered_map<string,int> map2;
        for(auto w1:words1){
            map1[w1]++;
        }
        for(auto w2:words2){
            map2[w2]++;
        }
        int res = 0;
        for(auto w1:words1){
            if(map1[w1] == 1 && map2[w1] == 1) res++;
        }
        return res;
    }
};
相关推荐
飞川撸码1 分钟前
【LeetCode 热题100】240:搜索二维矩阵 II(详细解析)(Go语言版)
leetcode·矩阵·golang
半盏茶香2 分钟前
启幕数据结构算法雅航新章,穿梭C++梦幻领域的探索之旅——堆的应用之堆排、Top-K问题
java·开发语言·数据结构·c++·python·算法·链表
小竹子1425 分钟前
L1-1 天梯赛座位分配
数据结构·c++·算法
董董灿是个攻城狮35 分钟前
Transformer 通关秘籍8:词向量如何表示近义词?
算法
独好紫罗兰1 小时前
洛谷题单2-P5712 【深基3.例4】Apples-python-流程图重构
开发语言·python·算法
uhakadotcom1 小时前
NVIDIA Resiliency Extension(NVRx)简介:提高PyTorch训练的容错性
算法·面试·github
梭七y1 小时前
【力扣hot100题】(020)搜索二维矩阵Ⅱ
算法·leetcode·职场和发展
v维焓2 小时前
C++(思维导图更新)
开发语言·c++·算法
ylfhpy2 小时前
Java面试黄金宝典22
java·开发语言·算法·面试·职场和发展
Phoebe鑫2 小时前
数据结构每日一题day9(顺序表)★★★★★
数据结构·算法