Question

面试题 17.13. 恢复空格

English Version

题目描述

哦，不！你不小心把一个长篇文章中的空格、标点都删掉了，并且大写也弄成了小写。像句子"I reset the computer. It still didn’t boot!"已经变成了"iresetthecomputeritstilldidntboot"。在处理标点符号和大小写之前，你得先把它断成词语。当然了，你有一本厚厚的词典dictionary，不过，有些词没在词典里。假设文章用sentence表示，设计一个算法，把文章断开，要求未识别的字符最少，返回未识别的字符数。

注意：本题相对原题稍作改动，只需返回未识别的字符数

 

示例：

输入：
dictionary = ["looked","just","like","her","brother"]
sentence = "jesslookedjustliketimherbrother"
输出： 7
解释： 断句后为"jess looked just like tim her brother"，共7个未识别字符。

提示：

• 0 <= len(sentence) <= 1000
• dictionary中总字符数不超过 150000。
• 你可以认为dictionary和sentence中只包含小写字母。

解法

方法一：动态规划

class Solution:
  def respace(self, dictionary: List[str], sentence: str) -> int:
    s = set(dictionary)
    n = len(sentence)
    dp = [0] * (n + 1)
    for i in range(1, n + 1):
      dp[i] = dp[i - 1] + 1
      for j in range(i):
        if sentence[j:i] in s:
          dp[i] = min(dp[i], dp[j])
    return dp[-1]

class Solution {
  public int respace(String[] dictionary, String sentence) {
    Set<String> dict = new HashSet<>(Arrays.asList(dictionary));
    int n = sentence.length();
    int[] dp = new int[n + 1];
    for (int i = 1; i <= n; i++) {
      dp[i] = dp[i - 1] + 1;
      for (int j = 0; j < i; ++j) {
        if (dict.contains(sentence.substring(j, i))) {
          dp[i] = Math.min(dp[i], dp[j]);
        }
      }
    }
    return dp[n];
  }
}

class Solution {
public:
  int respace(vector<string>& dictionary, string sentence) {
    unordered_set<string> s(dictionary.begin(), dictionary.end());
    int n = sentence.size();
    vector<int> dp(n + 1);
    for (int i = 1; i <= n; ++i) {
      dp[i] = dp[i - 1] + 1;
      for (int j = 0; j < i; ++j) {
        if (s.count(sentence.substr(j, i - j))) {
          dp[i] = min(dp[i], dp[j]);
        }
      }
    }
    return dp[n];
  }
};

func respace(dictionary []string, sentence string) int {
  s := map[string]bool{}
  for _, v := range dictionary {
    s[v] = true
  }
  n := len(sentence)
  dp := make([]int, n+1)
  for i := 1; i <= n; i++ {
    dp[i] = dp[i-1] + 1
    for j := 0; j < i; j++ {
      if s[sentence[j:i]] {
        dp[i] = min(dp[i], dp[j])
      }
    }
  }
  return dp[n]
}