leetcode30——串联所有单词的字串

Posted 2022-01-26 dtwd886

滑动窗口的思想，将所有子串分为单词长度个元组，作为遍历的外层循环。

内层循环跨度为单词的长度

class Solution 
    public List<Integer> findSubstring(String s, String[] words) 
        List<Integer>result=new ArrayList<Integer>();

        int wordNum=words.length;
        if(wordNum==0)
            return result;
        int wordLen=words[0].length();
        HashMap<String,Integer>hashMap1=new HashMap<String,Integer>();
        for(String word:words)
        
            if(hashMap1.containsKey(word))hashMap1.put(word,hashMap1.get(word)+1);
            else hashMap1.put(word,1);
        
        //分成wordLen类情况
        for(int j=0;j<wordLen;j++)
        
            HashMap<String,Integer>hashMap2=new HashMap<String,Integer>();
            int num=0;//记录当前HashMap2中有多少个单词与HashMap1匹配
            for(int i=j;i<s.length()-wordNum*wordLen+1;i+=wordLen)
            
                boolean hasRemoved=false;   //防止情况三移除后，情况一继续移除
                while(num<wordNum)
                    String word=s.substring(i+num*wordLen,i+(num+1)*wordLen);
                    if(hashMap1.containsKey(word))
                        int value=hashMap2.getOrDefault(word,0);
                        hashMap2.put(word,value+1);
                        if(hashMap2.get(word)>hashMap1.get(word))
                        
                            hasRemoved=true;
                            int removeNum=0;
                            while(hashMap2.get(word)>hashMap1.get(word))
                            
                                String curWord=s.substring(i+removeNum*wordLen,i+(removeNum+1)*wordLen);
                                int v=hashMap2.get(curWord);
                                hashMap2.put(curWord,v-1);
                                removeNum++;
                            
                            num=num-removeNum+1;      
                            i=i+(removeNum-1)*wordLen;
                            break;
                        
                    
                    else    //情况二，当前单词不在hashMap1
                    
                        hashMap2.clear();
                        i=i+num*wordLen;
                        num=0;
                        break;
                    
                    num++;
                
                if(num==wordNum) result.add(i);
                if(num==wordNum&&!hasRemoved)
                
                    String firstWord=s.substring(i,i+wordLen);
                    int v=hashMap2.get(firstWord);
                    hashMap2.put(firstWord,v-1);
                    num=num-1;
                
            
        
        return result;