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".
struct Solution;
impl Solution {
fn find_anagrams(s: String, p: String) -> Vec<i32> {
let mut res: Vec<i32> = vec![];
let mut ss: Vec<usize> = vec![0; 26];
let mut sp: Vec<usize> = vec![0; 26];
let s: Vec<u8> = s.bytes().collect();
let p: Vec<u8> = p.bytes().collect();
if s.len() < p.len() {
return res;
}
for i in 0..p.len() {
let c = s[i] as usize - 'a' as usize;
ss[c] += 1;
}
for i in 0..p.len() {
let c = p[i] as usize - 'a' as usize;
sp[c] += 1;
}
if ss == sp {
res.push(0);
}
for i in 1..=(s.len() - p.len()) {
let c = s[i - 1] as usize - 'a' as usize;
let d = s[..][i + p.len() - 1] as usize - 'a' as usize;
ss[c] -= 1;
ss[d] += 1;
if ss == sp {
res.push(i as i32);
}
}
res
}
}
#[test]
fn test() {
let s = "cbaebabacd".to_string();
let p = "abc".to_string();
assert_eq!(Solution::find_anagrams(s, p), vec![0, 6]);
let s = "abab".to_string();
let p = "ab".to_string();
assert_eq!(Solution::find_anagrams(s, p), vec![0, 1, 2]);
}