leetcode讲解--884. Uncommon Words from Two Sentences

最新推荐文章于 2025-09-05 11:17:42 发布

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最新推荐文章于 2025-09-05 11:17:42 发布

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文章标签：数据结构与算法 java

原文链接：https://segmentfault.com/a/1190000017577354

本文介绍了一种使用HashMap来找出两个句子中不常见单词的方法。通过遍历每个句子并将单词计数存入HashMap，可以有效地找出那些只在一个句子中出现一次的单词。最后，将这些单词收集到一个列表中并返回。

题目

We are given two sentences A and B. (A sentence is a string of space separated words. Each word consists only of lowercase letters.)

A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.

Return a list of all uncommon words.

You may return the list in any order.

Example 1:

Input: A = "this apple is sweet", B = "this apple is sour"
Output: ["sweet","sour"]

Example 2:

Input: A = "apple apple", B = "banana"
Output: ["banana"]

Note:

0 <= A.length <= 200
0 <= B.length <= 200
A and B both contain only spaces and lowercase letters.

题目地址

讲解

用hashmap记录每一个单词

Java代码

class Solution {
    private Map<String, Integer> map = new HashMap<>();
    public String[] uncommonFromSentences(String A, String B) {
        List<String> result = new ArrayList<>();
        countString(A);
        countString(B);
        for(String key:map.keySet()){
            if(map.get(key)==1){
                result.add(key);
            }
        }
        String[] str = new String[result.size()];
        return result.toArray(str);
    }
    
    private void countString(String A){
        int index=0;
        for(int i=0;i<A.length();i++){
            if(A.charAt(i)==' '){
                Integer count = map.get(A.substring(index, i));
                if(count==null){
                    map.put(A.substring(index, i), 1);
                }else{
                    map.put(A.substring(index, i), count+1);
                }
                if(i+1<A.length()){
                    index = i+1;
                }
            }else if(i==A.length()-1){
                Integer count = map.get(A.substring(index));
                if(count==null){
                    map.put(A.substring(index), 1);
                }else{
                    map.put(A.substring(index), count+1);
                }
            }
        }
    }
}