标签：att ice length lap none spl output 开头 counter
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".
结果有模板。几个大的步骤：end先移动 begin后移动
while(end < s.length()){ 从end开始减去s end++;//把map里的t在s里找一遍，移动end直到t结束 while(counter == 0){//找完了t,map已经空了的时候 从begin开头存一个s 存到结果集合 begin++;//begin往前一位 } }
test case:
"abab" "ab" stdout counter = 2 end = 0 结尾处的字母c = a counter--之后的mapsize = 1 end++之后 = 1 ------- end = 1 结尾处的字母c = b counter--之后的mapsize = 0 end++之后 = 2 ------- begin = 0 开头处的字母tempc = a //这个key还是有的 counter++之后的mapsize = 1 //就加了一个值 此时的begin = 0 end = 2 结尾处的字母c = a counter--之后的mapsize = 0 //这里还要减，所以下面加1 end++之后 = 3 //又找到了 ------- begin = 1 开头处的字母tempc = b counter++之后的mapsize = 1 此时的begin = 1 end = 3 结尾处的字母c = b counter--之后的mapsize = 0 end++之后 = 4 ------- begin = 2 开头处的字母tempc = a counter++之后的mapsize = 1 此时的begin = 2 counter形容的是0个值的key的个数
class Solution { public List<Integer> findAnagrams(String s, String t) { List<Integer> result = new ArrayList<Integer>(); HashMap<Character, Integer> map = new HashMap<>(); //cc if (s.length() < t.length()) return result; int begin, end = 0; //存t for (int i = 0; i < t.length(); i++) { char tChar = s.charAt(i); map.put(tChar, map.getOrDefault(tChar, 0) + 1); } int counter = map.size(); //从s结尾开始减，从s开头开始存 while (end < s.length()) { char cEnd = s.charAt(end); if (map.containsKey(cEnd)) { map.put(cEnd, map.get(cEnd) - 1); if (map.get(cEnd) == 0) counter--; } end++; //t里的都数完了 while (counter == 0) { char cBegin = s.charAt(begin); if (map.containsKey(cBegin)) { map.put(cBegin, map.get(cBegin) + 1); if (map.get(cBegin) > 0) counter++; } if (end - begin == t.length()) result.add(begin); begin++; } } return result; } }
438. Find All Anagrams in a String 438.查找字符串中的所有Anagrams
标签：att ice length lap none spl output 开头 counter
原文地址：https://www.cnblogs.com/immiao0319/p/13064168.html
