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Difficulty: Medium
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Related Topics: Hash Table
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Link: https://leetcode.com/problems/find-all-anagrams-in-a-string/
Description
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
给定一个字符串 s 和一个非空字符串 p，在 s 里寻找与 p 互为相同字母异序词的子串，返回这些子串的起始下标。
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
所有字符串只包含小写英文字母，s 和 p 的长度不超过 20,100。
The order of output does not matter.
可以以任意顺序输出结果。
Examples
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
对于相同字母异序词，首先想到的就是统计字母出现的频率。本题解法显而易见：先统计 p 的字频，然后再统计 s 从 0 开始的子串的字频，然后维护一个长度为 p.length 的滑动窗口，维护其中的字频并与目标值进行比较。考虑到 HashMap 有映射不存在的问题难以处理，本题的字符串又只包含小写英文字母，所以直接使用长度为 26 的数组代替 HashMap，代码如下：
class Solution {
fun findAnagrams(s: String, p: String): List<Int> {
// 没说 p 的长度一定比 s 小，所以需要判断
if (s.length < p.length) {
return emptyList()
}
val target = p.freqArray()
val cur = IntArray(26)
val result = arrayListOf<Int>()
// 第一趟
for (i in 0..p.lastIndex) {
cur[s[i] - 'a']++
}
if (cur.contentEquals(target)) {
result.add(0)
}
// 滑动窗口，每次往后移动一个字符，更新 cur 并与 target 比较
for (i in 1..(s.lastIndex - p.lastIndex)) {
cur[s[i - 1] - 'a']--
cur[s[i + p.lastIndex] - 'a']++
if (cur.contentEquals(target)) {
result.add(i)
}
}
return result
}
private fun String.freqArray(): IntArray {
val result = IntArray(26)
this.forEach { result[it - 'a']++ }
return result
}
}