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: 

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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: 

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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".

思路

思路很简单，利用两个哈希表/26元素的数组，记录下s, p字母的统计数目，从而可以得出结果。

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class Solution(object):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        res = []
        pdict = {}
        plen = len(p)
        slen = len(s)
        
        if(plen <= slen):
        
            for i in range(plen):
                if (pdict.has_key(p[i])):
                    pdict[p[i]] += 1
                else:
                    pdict[p[i]] = 1
                    
            #print pdict
            sdict = {}
            
            for n in range(plen):
                if (sdict.has_key(s[n])):
                    sdict[s[n]] += 1
                else:
                    sdict[s[n]] = 1
                    
            print sdict
            
            for m in range(slen-plen+1):
                flag = 1
                # 注意！在这个地方，"==","!="都是可以判断两个dict是否相等的操作符
                # 如果遍历字典，反而会使时间复杂度大大提升，从而得到TLE，得不偿失。
                if (sdict != pdict):
                    flag = 0
                if (flag):
                    res.append(m)
                if(m == slen-plen):
                    break      
                if (sdict.has_key(s[m]) and sdict[s[m]] != 1):
                    sdict[s[m]] -= 1
                else:
                    del sdict[s[m]]                
                if (sdict.has_key(s[m+plen])):
                    sdict[s[m+plen]] += 1
                else:
                    sdict[s[m+plen]] = 1
                    
        return res