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".
给一个字符串s和一个非空字符串p，找出s中所有的p的变位词的index。
解法：滑动窗口法，双指针。
Java:
class Solution {
public List<Integer> findAnagrams(String s, String p) {
int left = 0;
int right = 0;
int matchSize = p.length();
int[] map = new int[256];
List<Integer> res = new ArrayList<>();
// count the char number in p 计算p中各个字符的数量
for (char c:p.toCharArray()){
map[c] ++;
}
// build window 开始进行sliding window
while (right < s.length()){
// this char exists in p
// 如果当前的char是存在于p中，则目标大小matchsize就得减少
// 判断标准就是他的值不是为-1
if (map[s.charAt(right)] > 0)
matchSize --;
map[s.charAt(right)] --;
if (right - left == p.length()-1){
// check matchSize equals to 0
// 如果此时目标大小也是0，说明这就是需要的一个子串
if (matchSize == 0)
// add the left index
res.add(left);
// move left pointer to start new search
// 如果当这个字符原来是p中的话，现在移动指针需要还原以前原有的matchSize，开始新的搜索
if (map[s.charAt(left)] >= 0)
matchSize ++;
// 还原以前每个元素减去的1
map[s.charAt(left)]++;
left++;
}
right++;
}
return res;
}
}
Java:
public 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;
}
}
Python:
class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
result = []
cnts = [0] * 26
for c in p:
cnts[ord(c) - ord('a')] += 1
left, right = 0, 0
while right < len(s):
cnts[ord(s[right]) - ord('a')] -= 1
while left <= right and cnts[ord(s[right]) - ord('a')] < 0:
cnts[ord(s[left]) - ord('a')] += 1
left += 1
if right - left + 1 == len(p):
result.append(left)
right += 1
return result
Python: wo
class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
n = len(p)
counts = [0] * 26
for i in p:
counts[ord(i) - ord('a')] += 1
res = []
for i in xrange(len(s)):
counts[ord(s[i]) - ord('a')] -= 1
if i >= n:
counts[ord(s[i - n]) - ord('a')] += 1
if self.checkAllZero(counts):
res.append(i - n + 1)
return res
def checkAllZero(self, counts):
for i in counts:
if i != 0:
return False
return True
C++:
// Time: O(n)
// Space: O(1)
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> result;
if (p.empty() || s.empty()) {
return result;
}
vector<int> cnts(26);
for (const auto& c : p) {
++cnts[c - 'a'];
}
for (int left = 0, right = 0; right < s.length(); ++right) {
--cnts[s[right] - 'a'];
while (left <= right && cnts[s[right] - 'a'] < 0) {
++cnts[s[left++] - 'a'];
}
if (right - left + 1 == p.length()) {
result.emplace_back(left);
}
}
return result;
}
};
C++:
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
if (s.empty()) return {};
vector<int> res, m(256, 0);
int left = 0, right = 0, cnt = p.size(), n = s.size();
for (char c : p) ++m[c];
while (right < n) {
if (m[s[right++]]-- >= 1) --cnt;
if (cnt == 0) res.push_back(left);
if (right - left == p.size() && m[s[left++]]++ >= 0) ++cnt;
}
return res;
}
};
类似题目：
[LeetCode] 242. Valid Anagram 验证变位词
[LeetCode] 567. Permutation in String 字符串中的全排列