30. Substring with Concatenation of All Words

题目

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

For example, given:

s: “barfoothefoobarman”

words: [“foo”, “bar”]

You should return the indices: [0,9].

(order does not matter).

分析

解决该问题的关键是理解清楚要求。

给定一个目标字符串s，一个单词集合words。

要求使得words集合中所有元素连续出现在s中的首位置组成的集合（元素顺序不考虑）。

正如所给实例，目标字符串s: “barfoothefoobarman”

对比单词集合words: [“foo”, “bar”]

我们发现，在pos=0 ~ 5时“barfoo”恰好匹配，则0加入结果result；

在pos=9 ~ 14时“foobar”恰好匹配，则9加入结果result；

在理清楚题意后，便可入手程序实现。

class Solution(object):
##判断是否有words中的单组组成
def isEqual(self, s, words, n):
sList = []
i = 0
// 将字符串按照 words中每个单词的长度截取成数组
while(i<len(s)):
sList.append(s[i:i+n])
i = i + n
sList.sort()
words.sort()
if sList== words:
return True
else:
return False
def findSubstring(self, s, words):
"""
:type s: str
:type words: List[str]
:rtype: List[int]
"""
if len(words)<=0:
return
n = len(words[0])
lenWord = n *len(words)
result = []
i = 0
#注意while 的跳出条件 只需要循环 len(s)-lenWord次就够了
while(i<=len(s)-lenWord):
sCurrent = s[i:i+n]
if sCurrent in words:
#取出长度等于words中所有单词组成字符串的长度的S的字串
sSub = s[i:lenWord+i]
if self.isEqual(sSub, words,n):
result.append(i)
#每次i增1
i += 1
return result