Problem
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".
Solution
dumb version
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> res = new ArrayList<>();
if (s == null || p == null || s.length() < p.length()) return res;
for (int i = 0; i <= s.length()-p.length(); i++) {
if (isAnagram(s.substring(i, i+p.length()), p)) res.add(i);
}
return res;
}
private boolean isAnagram(String a, String b) {
int[] dict = new int[256];
for (char ch: a.toCharArray()) {
dict[ch]++;
}
for (char ch: b.toCharArray()) {
dict[ch]--;
if (dict[ch] < 0) return false;
}
return true;
}
}
sliding window
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> res = new LinkedList<>();
if (s.length() == 0 || p.length() == 0 || s.length() < p.length()) return res;
int[] dict = new int[26];
for (char ch: p.toCharArray()) {
dict[ch-'a']++;
}
int start = 0, end = p.length(), diff = p.length();
for (int i = 0; i < p.length(); i++) {
int index = s.charAt(i)-'a';
dict[index]--;
if (dict[index] >= 0) diff--;
}
if (diff == 0) res.add(0);
while (end < s.length()) {
int index = s.charAt(start)-'a';
if (dict[index] >= 0) diff++;
dict[index]++;
start++;
index = s.charAt(end)-'a';
dict[index]--;
if (dict[index] >= 0) diff--;
end++;
if (diff == 0) res.add(start);
}
return res;
}
}
update 2018-10 with comments
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> res = new LinkedList<>();
if (s.length() == 0 || p.length() == 0 || s.length() < p.length()) return res;
int[] dict = new int[26];
for (char ch: p.toCharArray()) {
dict[ch-'a']++;
}
int start = 0, end = p.length(), diff = p.length();
for (int i = 0; i < p.length(); i++) {
int index = s.charAt(i)-'a';
dict[index]--;
if (dict[index] >= 0) diff--;
}
if (diff == 0) res.add(0);
while (end < s.length()) {
//will release the s.charAt(start),
//so if >= 0 in map, that means we released one char in anagram, so increase diff by 1,
//else it's not in anagram of p, continue
//after release, increase it back in map
//shift start
int index = s.charAt(start)-'a';
if (dict[index] >= 0) diff++;
dict[index]++;
start++;
//will store the s.charAt(end),
//so first decrease in map,
//if still >= 0, that means we saved one char that is in anagram p, so reduce diff by 1
//else it's not in anagram of p, continue
//shift end
index = s.charAt(end)-'a';
dict[index]--;
if (dict[index] >= 0) diff--;
end++;
//if diff became 0 at the end of this iteration, add index start to result
if (diff == 0) res.add(start);
}
return res;
}
}
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