计算给定单词集中的子串答案

【问题标题】：Counting substrings from given set of words计算给定单词集中的子串
【发布时间】：2014-11-30 01:16:41
【问题描述】：

我有一组字符串（字典）和一个字符串 T，我必须计算我可以从字典中的单词构建 T 的次数

例如

字典包含： hello world llo he

和字符串 T "helloworld"

输出应该是 2，因为 "hellowold" 可以从 hello+world, he+llo+world 构建

有什么有效的算法可以做到这一点吗？

【问题讨论】：

标签： count substring

【解决方案1】：

这是一个在 python 中的快速实现：

from collections import defaultdict

def count_constructions( words, string ):
    # First we're going to make a map with 
    # positions mapped to lists of words starting
    # at that position in the string
    words_at_index = defaultdict( list )
    for word in words:    
        i = string.find(word)
        while i >= 0:
            words_at_index[i].append(word)
            i = string.find(word, i + 1)
    # I know there's a more pythonic way to do this, 
    # but the point here is to be able to inc count within
    # the auxilliary function
    count = [ 0 ]

    # This will find all of the ways to cover the remaining string
    # starting at start
    def recurse( start ):
        for w in words_at_index[start]:
            # w matches from string[start] to string[next_start]
            next_start = start + len(w)
            # see if we've covered the whole thing.        
            if next_start == len(string):
                count[0] += 1
                # we could also emit the words forming the string here
            else: 
                # otherwise, count the times we can cover it from
                # next_start on
                recurse(next_start)

    recurse(0)
    return count[0]


dictionary = [ 'hello', 'world', 'llo', 'he' ]
word = "helloworld"

print( count_constructions( dictionary, word ) )

【讨论】：

【解决方案2】：

我会首先从您的字典中获取一个子集，其中仅包含可能是您正在搜索的单词的一部分的单词。然后，剩下的话你可以做一个回溯实现，它不应该使用太多的资源，因为你将运行回溯的集合会非常小。

【讨论】：