leetcode 438. Find All Anagrams in a String和30. Substring with Concatenation of All Words

目录

438. 找到字符串中所有字母异位词

30. 串联所有单词的子串

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438. 找到字符串中所有字母异位词

滑动窗口+哈希表

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        int len_s = s.size();
        int len_p = p.size();
        if(len_s < len_p)
            return {};
        vector<int> p_table(26,0);
        vector<int> window_table(26,0);
        for(int i = 0;i < len_p;i++){
            p_table[p[i]-'a']++;
            window_table[s[i]-'a']++;
        }

        vector<int> res;
        if(p_table == window_table){
            res.push_back(0);
        }
        for(int left = 0;left < len_s - len_p;left++){
            window_table[s[left]-'a']--;
            window_table[s[left+len_p] -'a']++;
            if(window_table == p_table)
                res.push_back(left+1);
        }
        return res;
    }
};

30. 串联所有单词的子串

滑动窗口+哈希表

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        int s_len = s.size();
        int word_cnt = words.size();
        int word_len = words[0].size();
        int window_len = word_cnt*word_len;
        if(s_len < window_len)
            return {};

        unordered_map<string,int> words_table;
        for(int i = 0;i < word_cnt;i++){
            if(words_table.contains(words[i])){
                words_table[words[i]]++;
            }else{
                words_table[words[i]]=1;
            }
        }

        vector<int> res;

        unordered_map<string,int> window_table;

        //外层循环是对s做划分，例如：s = "barfoothefoobarman", words = ["foo","bar"]
        //此时word_len是3，所以有3种划分方式
        //对s从第一个字母开始划分为bar,foo,the,foo,bar,man
        //对s从第二个字母开始划分为b,arf,oot,hef,oob,arm,an。最前面的b和最后面的an不需要考虑
        //对s从第三个字母开始划分为ba,rfo,oth,efo,oba,rma,n。最前面的ba和最后面的n不需要考虑
        for(int start = 0; start< word_len && s_len - start >= window_len;start++){

            for(int i = 0;i < word_cnt;i++){
                string temp = s.substr(start + i*word_len,word_len);
                if(window_table.contains(temp))
                    window_table[temp]++;
                else
                    window_table[temp]=1;
            }
            if(check(window_table,words_table))
                res.push_back(start);

            int current_left = start;
            while(current_left < s_len - window_len){
                string temp = s.substr(current_left,word_len);
                window_table[temp]--;
                if(window_table[temp] == 0)
                    window_table.erase(temp);
                window_table[s.substr(current_left+window_len,word_len)]++;
                if(check(window_table,words_table))
                    res.push_back(current_left+word_len);
                current_left += word_len;
            }
            window_table.clear();
        }

        return res;
    }

    bool check(unordered_map<string,int> &window_table,unordered_map<string,int> &words_table){
        for(const auto&[word,cnt] : words_table){
            if(cnt != window_table[word])
                return false;
        }
        return true;
    }
};