438. Find All Anagrams in a String
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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".
第一次做这种题,想了很久都没有想出这种方法,用其它方法做的时候,时间又超了。
Sliding Window:大体思路就是用一个窗划过字符串s。窗内初始时装载目标字符串,窗划过字符串s,减去进入窗的字符,加上离开窗的字符。若窗内的字符数为0,则表明目标字符串依次进入了窗。
public class Solution {
public List<Integer> findAnagrams(String s, String p) {List<Integer> res = new ArrayList<>();
if(p.length() > s.length())
return res;
char[] sStr = s.toCharArray();
int[]map = new int[26];
for(char ch:p.toCharArray())
map[ch - 'a']++;
int j = 0;
for(j=0; j<p.length()-1; j++) //窗只出不进
map[sStr[j] - 'a']--;
for(int i=0; j<s.length(); i++, j++){
map[sStr[j] - 'a']--;
if(check(map))
res.add(i);
map[sStr[i] - 'a']++;
}
return res;
}
public boolean check(int[]map){
for(int n:map)
if(n > 0) return false;
return true;
}
}
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