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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

Python3

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]

Java

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];
    }
}

C++

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];
    }
};

Go

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]
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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