目錄

• 題目描述
• 題目大意
• 解題方法
• 滑動視窗
• 雙指標
• 日期

題目描述

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".

題目大意

在s中有多少個位置，從這個位置開始和p等長度的子字串中，所包含的字元和p是一樣的。

解題方法

滑動視窗

這個題考的是時間複雜度。如果判斷兩個切片是否是排列組合的話，時間複雜度略高，會超時。

能AC的做法是用了一個滑動視窗，每次進入視窗的字元的個數+1，超出滑動視窗的字元個數-1.

這樣就一遍就搞定了，而且不用每個切片都算是不是一個排列組合。

Counter大法好，判斷兩個字串是否是排列組合直接統計詞頻然後==判斷即可。

注意如果一個詞出現的次數是0，那麼需要從Counter中移除，因為Counter({‘a’: 0, ‘b’: 1}) w不等於Counter({‘b’: 1})。

from collections import Counter
class Solution(object
 
):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        answer = []
        m, n = len(s), len(p)
        if m < n:
            return answer
        pCounter = Counter(p)
        sCounter = Counter(s[:n-1])
        index = 0
        for index in xrange(n - 1, m):
                sCounter[s[index]] += 1
                if sCounter == pCounter:
                    answer.append(index - n + 1)
                sCounter[s[index - n + 1]] -= 1
                if sCounter[s[index - n + 1]] == 0:
                    del sCounter[s[index - n + 1]]
        return answer
            

雙指標

二刷的時候也是滑動視窗，但是使用的是雙指標的解法，看上去好像沒有上面這個方法更簡單。

class Solution(object):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        count = collections.Counter()
        M, N = len(s), len(p)
        left, right = 0, 0
        pcount = collections.Counter(p)
        res = []
        while right < M:
            count[s[right]] += 1
            if right - left + 1 == N:
                if count == pcount:
                    res.append(left)
                count[s[left]] -= 1
                if count[s[left]] == 0:
                    del count[s[left]]
                left += 1
            right += 1
        return res

日期

2018 年 1 月 27 日
2018 年 11 月 24 日 —— 週六快樂