438. Find All Anagrams in a String

题目

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.

題目大意

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

解題思路

跟 0567.Permutation-in-String類似, 只是把找到的答案記錄起來

來源

• https://leetcode.com/problems/find-all-anagrams-in-a-string/

• 解答

  https://github.com/kimi0230/LeetcodeGolang/blob/master/Leetcode/0438.Find-All-Anagrams-in-a-String/main.go

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
}