标签：-- while tar lock array code val ++ 字符
找到字符串中所有字母易位词。给定一个字符串 s 和一个非空字符串 p，找到 s 中所有是 p 的字母异位词的子串，返回这些子串的起始索引。例子，
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".
思路是滑动窗口（sliding window）。这个题也可以用之前的模板做，参见76，159，340。先遍历p，得到p中每个字母和他们对应的出现次数。再去遍历s，用start和end指针，end在前，当某个字母的value位0的时候，counter才能--。当counter == 0的时候，判断end - begin是否等于p的长度，若是，则将begin的坐标加入结果集。
时间O(n)
空间O(1)
Java实现
1 class Solution { 2 public List<Integer> findAnagrams(String s, String p) { 3 List<Integer> res = new ArrayList<>(); 4 // corner case 5 if (p.length() > s.length()) { 6 return res; 7 } 8 9 // normal case 10 Map<Character, Integer> map = new HashMap<>(); 11 for (char c : p.toCharArray()) { 12 map.put(c, map.getOrDefault(c, 0) + 1); 13 } 14 int counter = map.size(); 15 int begin = 0; 16 int end = 0; 17 while (end < s.length()) { 18 char c = s.charAt(end); 19 if (map.containsKey(c)) { 20 map.put(c, map.get(c) - 1); 21 if (map.get(c) == 0) { 22 counter--; 23 } 24 } 25 end++; 26 27 while (counter == 0) { 28 char temp = s.charAt(begin); 29 if (map.containsKey(temp)) { 30 map.put(temp, map.get(temp) + 1); 31 if (map.get(temp) > 0) { 32 counter++; 33 } 34 } 35 if (end - begin == p.length()) { 36 res.add(begin); 37 } 38 begin++; 39 } 40 } 41 return res; 42 } 43 }
JavaScript实现
[LeetCode] 438. Find All Anagrams in a String
标签：-- while tar lock array code val ++ 字符
原文地址：https://www.cnblogs.com/aaronliu1991/p/12626540.html
