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".
Approach #1: C++. Using sort and substr[Time Limit Exceeded]
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> ans;
if (s.length() < p.length()) return ans;
sort(p.begin(), p.end());
int len = p.size();
for (int i = 0; i < s.length()-len+1; ++i) {
string temp = s.substr(i, len);
sort(temp.begin(), temp.end());
if (temp == p) ans.push_back(i);
}
return ans;
}
};
Approach #2: C++. Using sliding windows.
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
unordered_map<char, int> mp;
vector<int> ans;
for (int i = 0; i < p.length(); ++i) {
mp[p[i]]++;
}
int begin = 0, end = 0;
int counter = mp.size();
while (end < s.length()) {
char tempc = s[end];
if (mp.count(tempc)) {
mp[tempc]--;
if (mp[tempc] == 0) counter--;
}
end++;
while (counter == 0) {
char tempc = s[begin];
if (mp.count(tempc)) {
mp[tempc]++;
if (mp[tempc] > 0) counter++;
}
if (end - begin == p.length()) {
ans.push_back(begin);
}
begin++;
}
}
return ans;
}
};
Approach #3: Java.
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++;
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;
}
}
Approach #4: Python.
from collections import Counter
class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
res = []
pCounter = Counter(p)
sCounter = Counter(s[:len(p)-1])
for i in range(len(p)-1, len(s)):
sCounter[s[i]] += 1
if sCounter == pCounter:
res.append(i-len(p)+1)
sCounter[s[i-len(p)+1]] -= 1
if sCounter[s[i-len(p)+1]] == 0:
del sCounter[s[i-len(p)+1]]
return res
| Time Submitted | Status | Runtime | Language |
|---|---|---|---|
| a few seconds ago | Accepted | 240 ms | python |
| 43 minutes ago | Accepted | 44 ms | cpp |
| an hour ago | Accepted | 61 ms | java |
438. Find All Anagrams in a String
原文：https://www.cnblogs.com/ruruozhenhao/p/9971035.html