给定字符串S和一组单词D，找到D中作为S的子序列的最长单词

Question

我尝试使用正则表达式在python 2.7中解决这个问题。我发现它很简单。想知道我是否遗漏了什么。这里有边缘表壳或表壳吗？

S = "abppplee"
D = {"able", "ale", "apple", "bale", "kangaroo"}

def subseq(S,D):  
    # every word in D is made into a regex pattern interspersed with .* and looked up in S
    import re
    in_dict = {word:len(word) for word in D if bool(re.search(pattern=".*".join(word),string=S))}
    return max(in_dict, key=in_dict.get)

subseq(S,D)

返回'apple'

Answer 1

的 re.search

扫描字符串以查找与模式匹配的内容，返回匹配对象，如果未找到匹配项则扫描无。

但在你的情况下，你需要找到带有列表的字符串的子序列。 生成子序列的代码是

def subsequence(X, Y, m, n):
    L = [[0 for x in range(n + 1)] for x in range(m + 1)]

    # Following steps build L[m+1][n+1] in bottom up fashion. Note
    # that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
    for i in range(m + 1):
        for j in range(n + 1):
            if i == 0 or j == 0:
                L[i][j] = 0
            elif X[i - 1] == Y[j - 1]:
                L[i][j] = L[i - 1][j - 1] + 1
            else:
                L[i][j] = max(L[i - 1][j], L[i][j - 1])

    # Following code is used to print LCS
    val = L[m][n]
    index = L[m][n]

    # Create a character array to store the lcs string
    lcs = [""] * (index + 1)
    lcs[index] = "\0"

    # Start from the right-most-bottom-most corner and
    # one by one store characters in lcs[]
    i = m
    j = n
    while i > 0 and j > 0:

        # If current character in X[] and Y are same, then
        # current character is part of LCS
        if X[i - 1] == Y[j - 1]:
            lcs[index - 1] = X[i - 1]
            i -= 1
            j -= 1
            index -= 1

        # If not same, then find the larger of two and
        # go in the direction of larger value
        elif L[i - 1][j] > L[i][j - 1]:
            i -= 1
        else:
            j -= 1

    sequence = "".join(lcs)
    return val,sequence

#Driver Code
S = "alnegearoo"
D = {"able", "ale", "apple", "bale", "kangaroo"}
max_len = 0
max_string = ""
for word in D:
    subseq_len,match = subsequence(word,S,len(word),len(S))
    if(max_len < subseq_len):
        max_len = subseq_len
        max_string = match

print(max_string)

输出

angaroo

享受编码......

Answer 2

我是用java做的，希望它能帮助别人解决它

import React, { Component } from 'react';
import {
  Text,
  View,
  ScrollView
} from 'react-native';

const text = ".\n".repeat(600)

export default class App extends Component {
  render() {
    return (
      <View style={{
        flex: 1,
        justifyContent: 'flex-start',
        alignItems: 'center',
        backgroundColor: '#F5FCFF',
      }}>
        <ScrollView style={{flex: 1}}>
          <Text>{text}</Text>
        </ScrollView>
      </View>
    );
  }
}

Answer 3

您应该使用re.escape。

import re
string = 'abppplee{lkfj3lkj.f*)(*)(*@}dddd.sdf*3j3jj3}@@@@'
words = {'apple', 'ale', 'bale', 'able', 'apple{}.*', 'kangaroo'}
max({word for word in words
     if re.search('.*'.join(map(re.escape, word)), string)},
    key=len)

我很难区分回答这个问题和例如做Kata。 Codewars。