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;
}
};
reference:
stackoverflow:what is sliding window algorithm
github:chaoyanghe silding window algortihm template