Question

我有一个字符串S（带有从0索引的单词）和一个子串Q.我希望在S中找到包含Q中所有单词的最小范围[L，R] .Q中没有重复的单词。我接近这个吗？

例如，

输入： S ：那个懒惰的棕色狐狸怎么样跳过另一只棕色的狐狸，懒狗吃了狐狸的食物 Q ：懒惰的棕色狗

输出： [11,15]

我的代码：

S = raw_input().strip().split(' ')
Q = raw_input().strip().split(' ')

count = [0 for x in range(len(Q))]
smallest_index = [0 for x in range(len(Q))]
largest_index = [0 for x in range(len(Q))]

for i in range(len(S)):
    for j in range(len(Q)):
        if S[i] == Q[j]:
            count[j] += 1
            if count[j] <= 1:
                smallest_index[j] = i
                largest_index[j] = i
            if count[j] > 1:
                largest_index[j] = i

largest_index.sort()
print "[%d," % largest_index[0],
print "%d]" % largest_index[len(Q)-1]

Answer 1

此代码并不特别有效，但它确实可以正常工作。也许有人会设计出比使用product更好的处理位置信息的方法。与此同时，您可以使用此代码来测试其他算法。

from itertools import product

def words_range(src, query):
    # Create a dict to store the word positions in src of each query word
    pos = {s: [] for s in query}
    for i, s in enumerate(src):
        if s in pos:
            pos[s].append(i)
    print(pos)

    # Find all the ranges that hold all the query word 
    ranges = ((min(t), max(t)) for t in product(*pos.values()))
    # Find the smallest range
    return min(ranges, key=lambda t:t[1] - t[0])

# Test

src = '''what about the lazy brown fox that jumped over the other
brown one which lazy dog ate the food of the fox'''.split()
for i, s in enumerate(src):
    print(i, s)

query = 'lazy brown dog'.split()
print(words_range(src, query))

query = 'the lazy brown fox'.split()
print(words_range(src, query))

<强>输出

0 what
1 about
2 the
3 lazy
4 brown
5 fox
6 that
7 jumped
8 over
9 the
10 other
11 brown
12 one
13 which
14 lazy
15 dog
16 ate
17 the
18 food
19 of
20 the
21 fox
{'lazy': [3, 14], 'brown': [4, 11], 'dog': [15]}
(11, 15)
{'the': [2, 9, 17, 20], 'lazy': [3, 14], 'brown': [4, 11], 'fox': [5, 21]}
(2, 5)

Answer 2

这是PM 2Ring's solution稍高效的版本，用循环取代了对from itertools import product def words_range(src, query): query = set(query) # Create a dict to store the word positions in src of each query word pos = {s: [] for s in query} for i, s in enumerate(src): if s in pos: pos[s].append(i) # Find all the ranges that hold all the query word # We'll iterate over the input string and keep track of # where each word appeared last last_pos = {} ranges = [] for i, word in enumerate(src): if word in query: last_pos[word] = i if len(last_pos) == len(query): ranges.append( (min(last_pos.values()), i) ) # Find the smallest range return min(ranges, key=lambda t:t[1] - t[0])的调用：

min(last_pos.values())

它不是非常线性的时间（因为循环中的min），但它是朝着正确方向迈出的一步。可能有一种方法可以摆脱v2_soap调用（我现在无法想到），这会使这种线性化。

Answer 3

这是基于@PM 2Ring答案的另一种方法：

S ='what about the lazy brown fox that jumped over the other brown one which lazy dog ate the food of the fox'
Q ='lazy brown dog'

import itertools
track={}

for index,value in enumerate(S.split()):

    if value in Q:
        if value not in track:
            track[value]=[index]
        else:
            track[value].append(index)


combination = [(min(item),max(item)) for item in itertools.product(*track.values())]


result=min([(i[1]-i[0],(i[0],i[1])) for i in combination if set(Q.split()).issubset(S.split()[i[0]:i[1]+1])])
print(result[1])

输出：

(11, 15)

字符串中最小范围的子字符串

3 个答案: