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。
class Solution {
public int respace(String[] dictionary, String sentence) {
int n = sentence.length();
Trie root = new Trie();
for (String word: dictionary) {
root.insert(word);
}
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
for (int i = 1; i <= n; ++i) {
dp[i] = dp[i - 1] + 1;
Trie curPos = root;
for (int j = i; j >= 1; --j) {
int t = sentence.charAt(j - 1) - 'a';
if (curPos.next[t] == null) {
break;
} else if (curPos.next[t].isEnd) {
dp[i] = Math.min(dp[i], dp[j - 1]);
}
if (dp[i] == 0) {
break;
}
curPos = curPos.next[t];
}
}
return dp[n];
}
}
class Trie {
public Trie[] next;
public boolean isEnd;
public Trie() {
next = new Trie[26];
isEnd = false;
}
public void insert(String s) {
Trie curPos = this;
for (int i = s.length() - 1; i >= 0; --i) {
int t = s.charAt(i) - 'a';
if (curPos.next[t] == null) {
curPos.next[t] = new Trie();
}
curPos = curPos.next[t];
}
curPos.isEnd = true;
}
}