| title | tags | author | description |
|---|---|---|---|
0438.Find All Anagrams in a String |
Medium, Sliding Window |
Kimi Tsai <[email protected]> |
Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
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".
Constraints:
1 <= s.length, p.length <= 3 * 104 s and p consist of lowercase English letters.
找所有字母異位詞, 就像全排列 給定一個字符串 S 和非空的字符串 P, 找到 S 中所有是 P 得排列, 並返回他的起始 index
跟 0567.Permutation-in-String類似, 只是把找到的答案記錄起來
package findallanagramsinastring
func FindAnagrams(s string, p string) []int {
need, window := make(map[rune]int), make(map[rune]int)
for _, c := range p {
need[c]++
}
left, right := 0, 0
valid := 0
res := []int{} //紀錄結果
for right < len(s) {
c := rune(s[right])
right++
if need[c] > 0 {
window[c]++
if window[c] == need[c] {
valid++
}
}
// fmt.Printf("[%d,%d) \n", left, right)
// 判斷左視窗是否收縮, 看看視窗長度是否同要找的字串的長度
if (right - left) >= len(p) {
if valid == len(need) {
// 想要的字元都找到了, 紀錄index
res = append(res, left)
}
d := rune(s[left])
left++
if need[d] > 0 {
if window[d] == need[d] {
valid--
}
window[d]--
}
}
}
return res
}
// 用 slice 取代 map 來優化
func FindAnagramsSlice(s string, p string) []int {
need := [256]int{}
for _, c := range p {
need[c-'a']++
}
left, right := 0, 0
count := len(p)
res := []int{}
for right < len(s) {
c := s[right] - 'a'
if need[c] > 0 {
count--
}
need[c]--
right++
if count == 0 {
res = append(res, left)
}
if (right - left) >= len(p) {
d := s[left] - 'a'
if need[d] >= 0 {
count++
}
need[d]++
left++
}
}
return res
}