[leetcode]Substring with Concatenation of All Words @ Python

原题地址：https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/

题意：

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S:

"barfoothefoobarman"

L:

["foo", "bar"]

You should return the indices:

[0,9]

.
(order does not matter).

解题思路：使用一个字典统计一下L中每个单词的数量。由于每个单词的长度一样，以题中给的例子而言，可以3个字母3个字母的检查，如果不在字典中，则break出循环。有一个技巧是建立一个临时字典curr，用来统计S中那些在L中的单词的数量，必须和L中单词的数量相等，否则同样break。

代码：

class Solution:
# @param S, a string
# @param L, a list of string
# @return a list of integer
def findSubstring(self, S, L):
words={}
wordNum=len(L)
for i in L:
if i not in words:
words[i]=1
else:
words[i]+=1
wordLen=len(L[0])
res=[]
for i in range(len(S)+1-wordLen*wordNum):
curr={}; j=0
while j<wordNum:
word=S[i+j*wordLen:i+j*wordLen+wordLen]
if word not in words:
break
if word not in curr:
curr[word]=1
else:
curr[word]+=1
if curr[word]>words[word]: break
j+=1
if j==wordNum: res.append(i)
return res