ST算法 Sliding Window algorithm template

ST算法(Sliding Window):A easy way to slove the substring problems

algorithm template to slove substring search problems

这是一个利用哈希思想解决字符串问题，特别是关于子字符串的问题的模版

方法描述: 怎样实现滑动窗口

Generally speaking a sliding window is a sub-list that runs over an underlying collection. I.e., if you have an array like

[a b c d e f g h]

a sliding window of size 3 would run over it like

[a b c]
  [b c d]
    [c d e]
      [d e f]
        [e f g]
          [f g h]

implement the template with java

public class Solution {
    public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) {
        //init a collection or int value to save the result according the question.
        List<Integer> result = new LinkedList<>();
        if(t.length()> s.length()) return result;
        
        //create a hashmap to save the Characters of the target substring.
        //(K, V) = (Character, Frequence of the Characters)
        Map<Character, Integer> map = new HashMap<>();
        for(char c : t.toCharArray()){
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        //maintain a counter to check whether match the target string.
        int counter = map.size();//must be the map size, NOT the string size because the char may be duplicate.
        
        //Two Pointers: begin - left pointer of the window; end - right pointer of the window
        int begin = 0, end = 0;
        
        //the length of the substring which match the target string.
        int len = Integer.MAX_VALUE; 
        
        //loop at the begining of the source string
        while(end < s.length()){
            
            char c = s.charAt(end);//get a character
            
            if( map.containsKey(c) ){
                map.put(c, map.get(c)-1);// plus or minus one
                if(map.get(c) == 0) counter--;//modify the counter according the requirement(different condition).
            }
            end++;
            
            //increase begin pointer to make it invalid/valid again
            while(counter == 0 /* counter condition. different question may have different condition */){
                
                char tempc = s.charAt(begin);//***be careful here: choose the char at begin pointer, NOT the end pointer
                if(map.containsKey(tempc)){
                    map.put(tempc, map.get(tempc) + 1);//plus or minus one
                    if(map.get(tempc) > 0) counter++;//modify the counter according the requirement(different condition).
                }
                
                /* save / update(min/max) the result if find a target*/
                // result collections or result int value
                
                begin++;
            }
        }
        return result;
    }
}

实现滑动窗口算法的c++模版

class Solution{
public:
    vector<int> slidingWindowTemplate(string s,string p){
        // define a vector store the result position
        vector<int> result;
        // string not have anagram
        if(s.size()<p.size())
            return result;
        map<char,int> hash_map;
        //  hash map string p 
        for(auto &c:p)
            map[p]++;
        // use counter store the hash map size
        int counter=map.size();
        // define two pointer,point to the window boundary
        int left=0,right=0;
        // the substring length
        int len=INT_MAX;
        for(right<s.size()){
            char c=s[right];
            if (hash_map.contains(c)){
                hash_map[c]--;
                if(hash_map[c]==0)
                    counter--;
            }
            right++;
            while(counter==0){
                char temp_c=s[left];
                if(hash_map.cotains(c)){
                    map[temp_c]++;
                    if(hash_map[temp_c]>0)
                        counter++;
                }
                left++;
            }
        }
        return result;


    }
}

最容易理解的版本

// 解决实际问题
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> pv(26,0), sv(26,0), res;
        if(s.size() < p.size())
           return res;
        // fill pv, vector of counters for pattern string and sv, vector of counters for the sliding window
        for(int i = 0; i < p.size(); ++i)
        {
            ++pv[p[i]-'a'];
            ++sv[s[i]-'a'];
        }
        if(pv == sv)
           res.push_back(0);

        //here window is moving from left to right across the string. 
        //window size is p.size(), so s.size()-p.size() moves are made 
        for(int i = p.size(); i < s.size(); ++i) 
        {
             // window extends one step to the right. counter for s[i] is incremented 
            ++sv[s[i]-'a'];
            
            // since we added one element to the right, 
            // one element to the left should be forgotten. 
            //counter for s[i-p.size()] is decremented
            --sv[s[i-p.size()]-'a']; 

            // if after move to the right the anagram can be composed, 
            // add new position of window's left point to the result 
            if(pv == sv)  
               res.push_back(i-p.size()+1);
        }
        return res;
    }
};