[LeetCode]Sliding Window Algorithm相关题目总结【重要】

438. Find All Anagrams in a String

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:s: "cbaebabacd" p: "abc"Output:[0, 6]Explanation:The substring with start index = 0 is "cba", which is an anagram of "abc".The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:s: "abab" p: "ab"Output:[0, 1, 2]Explanation:The substring with start index = 0 is "ab", which is an anagram of "ab".The substring with start index = 1 is "ba", which is an anagram of "ab".The substring with start index = 2 is "ab", which is an anagram of "ab".

一开始我用了一种O(N^2)的方法，超时。如下：

public class Solution {    public List<Integer> findAnagrams(String s, String p) {        if(p.length()>s.length() || s==null || p==null) return new ArrayList();                List<Character> list = new LinkedList<>();        List<Integer> res = new ArrayList<>();        int n=p.length();                        for(int i=0; i<s.length()-n+1; i++){            boolean find = true;            for(int k=0; k<n; k++){                list.add(p.charAt(k));            }                    for(int j=i; j<i+n; j++){                if(!list.contains(s.charAt(j)))                    {find=false;break;}                else{                    int index = list.indexOf(s.charAt(j));                    list.remove(index);                }            }                        list.clear();            if(find)                res.add(i);        }                return res;    }}

用滑动窗口算法的话，类似这种的substring search problem，都可以用O(N)的时间复杂度解决掉。

算法的模版如下：

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;    }}

该题的解法：

public class Solution {    public List<Integer> findAnagrams(String s, String p) {        List<Integer> res = new ArrayList<>();        if(p.length()>s.length()) return res;                HashMap<Character, Integer> map = new HashMap<>();        for(char c: p.toCharArray())            map.put(c,map.getOrDefault(c,0)+1);        int diff = map.size();        int begin=0, end=0;                while(end<s.length()){            char cur = s.charAt(end);            if(map.containsKey(cur)){                map.put(cur,map.get(cur)-1);                if(map.get(cur)==0)                    diff--;            }            end++;                        while(diff==0){                if(end-begin==p.length())                    res.add(begin);                                char temp = s.charAt(begin);                if(map.containsKey(temp)){                    map.put(temp, map.get(temp)+1);                    if(map.get(temp)>0)                        diff++;                }                begin++;            }        }                return res;    }}

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