438 Find All Anagrams in a String 找到字符串中所有字母异位词
Description:
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:
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".
题目描述:
给定一个字符串 s 和一个非空字符串 p，找到 s 中所有是 p 的字母异位词的子串，返回这些子串的起始索引。
字符串只包含小写英文字母，并且字符串 s 和 p 的长度都不超过 20100。
说明：
字母异位词指字母相同，但排列不同的字符串。
不考虑答案输出的顺序。
示例：
示例 1:
输入:
s: "cbaebabacd" p: "abc"
输出:
[0, 6]
解释:
起始索引等于 0 的子串是 "cba", 它是 "abc" 的字母异位词。
起始索引等于 6 的子串是 "bac", 它是 "abc" 的字母异位词。
示例 2:
输入:
s: "abab" p: "ab"
输出:
[0, 1, 2]
解释:
起始索引等于 0 的子串是 "ab", 它是 "ab" 的字母异位词。
起始索引等于 1 的子串是 "ba", 它是 "ab" 的字母异位词。
起始索引等于 2 的子串是 "ab", 它是 "ab" 的字母异位词。
思路:
滑动窗口
- 固定窗口大小为 p的长度, 每次移动窗口比较大小, 满足要求就加入结果
- 不固定窗口长度, 每次移动窗口比较 p是是否在窗口中, 满足则比较长度, 长度也符合的加入结果
时间复杂度O(n * m), 空间复杂度O(m), n为字符串 s的长度, m为字符串 p的长度
代码:
C++:
class Solution
{
public:
vector<int> findAnagrams(string s, string p)
{
vector<int> result;
int left = 0, right = 0, match = 0;
unordered_map<char, int> count, window;
for (auto c : p) ++count[c];
while (right < s.size())
{
char c1 = s[right];
if (count.count(c1))
{
window[c1]++;
if (window[c1] == count[c1]) match++;
}
right++;
while (match == count.size())
{
if (right - left == p.size()) result.emplace_back(left);
char c2 = s[left];
if (count.count(c2))
{
window[c2]--;
if (window[c2] < count[c2]) match--;
}
left++;
}
}
return result;
}
};
Java:
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> result = new ArrayList<>();
if (s.length() < p.length()) return result;
int[] array_s = new int[26], array_p = new int[26];
for (int i = 0; i < p.length(); i++) {
array_p[p.charAt(i) - 'a']++;
array_s[s.charAt(i) - 'a']++;
}
array_s[s.charAt(p.length() - 1) - 'a']--;
for (int i = p.length() - 1; i < s.length(); i++) {
array_s[s.charAt(i) - 'a']++;
if (isSame(array_p, array_s)) result.add(i - p.length() + 1);
array_s[s.charAt(i - p.length() + 1) - 'a']--;
}
return result;
}
private boolean isSame(int[] a, int[] b) {
for (int i = 0; i < 26; i++) {
if (a[i] != b[i]) return false;
}
return true;
}
}
Python:
class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
result = []
if len(p) > len(s):
return result
array_s, array_p = [0] * 26, [0] * 26
left, right = 0, len(p) - 2
for i in range(len(p)):
array_p[ord(p[i]) - ord('a')] += 1
array_s[ord(s[i]) - ord('a')] += 1
array_s[ord(s[len(p) - 1]) - ord('a')] -= 1
while right < len(s) - 1:
right += 1
array_s[ord(s[right]) - ord('a')] += 1
if array_p == array_s:
result.append(left)
array_s[ord(s[left]) - ord('a')] -= 1
left += 1
return result
