查看原文,請點這個連接java
我發如今leetcode的問題中,至少有5個子字符串尋找問題能夠用滑動窗口算法解決,所以我在這裏總結了這類算法的模版,但願能夠幫助你。算法
public class Solution { public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) { //根據問題,初始化一個儲存結果的容器 List<Integer> result = new LinkedList<>(); if (t.length()> s.length()) return result; //建立一個hashmap來保存目標子串中的字符 //(K, V) = (Character, Frequence of the Characters) //key是字符, value是該字符出現的次數 Map<Character, Integer> map = new HashMap<>(); //將目標子串轉爲map存儲 for(char c : t.toCharArray()){ map.put(c, map.getOrDefault(c, 0) + 1); } //維護一個計數器,去檢查是否匹配目標字符串 int counter = map.size();//必須是map的長度,不是字符串的長度是由於可能元素有重複。 //兩個點,窗口的左端點和右端點 int begin = 0, end = 0; //匹配目標字符串的子字符串的長度 int len = Integer.MAX_VALUE; //從源字符串循環 while (end < s.length()) { char c = s.charAt(end);//獲得右端點處的字符 if (map.containsKey(c)) { map.put(c, map.get(c)-1);//加一或減一 if (map.get(c) == 0) counter--;//根據不一樣的條件修改計數器 } end++; //increase begin pointer to make it invalid/valid again //計數器條件:不一樣的問題選擇不一樣的條件 while (counter == 0) { char tempc = s.charAt(begin);//注意:選擇字符是在開始端點而不是結束端點 if (map.containsKey(tempc)) { map.put(tempc, map.get(tempc) + 1);//加減一 if (map.get(tempc) > 0) counter++;//根據不一樣的需求修改計數器 } /* save / update(min/max) the result if find a target*/ //若是發現一個目標,保存、更新(最小、最大)結果 // result collections or result int value begin++; } } return result; } }
https://leetcode-cn.com/probl...
https://leetcode-cn.com/probl...
https://leetcode-cn.com/probl...
https://leetcode-cn.com/probl...
https://leetcode-cn.com/probl...code
438.找到字符串中全部字母異位詞
https://leetcode-cn.com/probl...leetcode
public class Solution { public List<Integer> findAnagrams(String s, String t) { List<Integer> result = new LinkedList<>(); if(t.length()> s.length()) return result; Map<Character, Integer> map = new HashMap<>(); for(char c : t.toCharArray()){ map.put(c, map.getOrDefault(c, 0) + 1); } int counter = map.size(); int begin = 0, end = 0; int head = 0; int len = Integer.MAX_VALUE; while (end < s.length()) { char c = s.charAt(end); if (map.containsKey(c)) { map.put(c, map.get(c) - 1); if (map.get(c) == 0) counter--; } end++; //counter等於0意味着,end以前至少有可以湊出target的字母數量 while (counter == 0) { char tempc = s.charAt(begin); if (map.containsKey(tempc)) { map.put(tempc, map.get(tempc) + 1); if(map.get(tempc) > 0){ counter++; } } if (end - begin == t.length()) { result.add(begin); } begin++; } } return result; } }
76. 最小覆蓋子串rem
https://leetcode-cn.com/probl...字符串
public class Solution { public String minWindow(String s, String t) { if(t.length()> s.length()) return ""; Map<Character, Integer> map = new HashMap<>(); for(char c : t.toCharArray()){ map.put(c, map.getOrDefault(c,0) + 1); } int counter = map.size(); int begin = 0, end = 0; int head = 0; int len = Integer.MAX_VALUE; while(end < s.length()){ char c = s.charAt(end); if( map.containsKey(c) ){ map.put(c, map.get(c)-1); if(map.get(c) == 0) counter--; } end++; while(counter == 0){ char tempc = s.charAt(begin); if(map.containsKey(tempc)){ map.put(tempc, map.get(tempc) + 1); if(map.get(tempc) > 0){ counter++; } } if(end-begin < len){ len = end - begin; head = begin; } begin++; } } if(len == Integer.MAX_VALUE) return ""; return s.substring(head, head+len); } }
3.無重複字符的最長子串get
https://leetcode-cn.com/probl...string
public class Solution { public int lengthOfLongestSubstring(String s) { Map<Character, Integer> map = new HashMap<>(); int begin = 0, end = 0, counter = 0, d = 0; while (end < s.length()) { // > 0 means repeating character //if(map[s.charAt(end++)]-- > 0) counter++; char c = s.charAt(end); map.put(c, map.getOrDefault(c, 0) + 1); if(map.get(c) > 1) counter++; end++; while (counter > 0) { //if (map[s.charAt(begin++)]-- > 1) counter--; char charTemp = s.charAt(begin); if (map.get(charTemp) > 1) counter--; map.put(charTemp, map.get(charTemp)-1); begin++; } d = Math.max(d, end - begin); } return d; } }
30.串聯全部單詞的子串hash
https://leetcode-cn.com/probl...it
public class Solution { public List<Integer> findSubstring(String S, String[] L) { List<Integer> res = new LinkedList<>(); if (L.length == 0 || S.length() < L.length * L[0].length()) return res; int N = S.length(); int M = L.length; // *** length int wl = L[0].length(); Map<String, Integer> map = new HashMap<>(), curMap = new HashMap<>(); for (String s : L) { if (map.containsKey(s)) map.put(s, map.get(s) + 1); else map.put(s, 1); } String str = null, tmp = null; for (int i = 0; i < wl; i++) { int count = 0; // remark: reset count int start = i; for (int r = i; r + wl <= N; r += wl) { str = S.substring(r, r + wl); if (map.containsKey(str)) { if (curMap.containsKey(str)) curMap.put(str, curMap.get(str) + 1); else curMap.put(str, 1); if (curMap.get(str) <= map.get(str)) count++; while (curMap.get(str) > map.get(str)) { tmp = S.substring(start, start + wl); curMap.put(tmp, curMap.get(tmp) - 1); start += wl; //the same as https://leetcode.com/problems/longest-substring-without-repeating-characters/ if (curMap.get(tmp) < map.get(tmp)) count--; } if (count == M) { res.add(start); tmp = S.substring(start, start + wl); curMap.put(tmp, curMap.get(tmp) - 1); start += wl; count--; } }else { curMap.clear(); count = 0; start = r + wl;//not contain, so move the start } } curMap.clear(); } return res; } }