[LeetCode] Word Break II

最新推荐文章于 2025-09-05 14:42:49 发布

转载最新推荐文章于 2025-09-05 14:42:49 发布 · 53 阅读

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4565214.html

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#开发工具

本文介绍了一种有效避免TLE问题的Word Break算法实现。通过预先判断字符串是否可由字典中的单词组成，并采用回溯法进行单词拆分，确保了算法效率。代码中使用了动态规划判断字符串是否可被拆分，以及回溯法进行实际的单词拆分。

Well, it seems that many people meet the TLE problem. Well, I use a simple trick in my code to aoivd TLE. That is, each time before I try to break s, I call isBreakable (just wordBreak from Word Break) to see whether it is breakable. If it is not breakable, then we need not move on. Otherwise, we break it using backtracking.

The idea of backtracking is also typical. Start from index 0 of s, each time we see a word that is in wordDict, add it to a temporary string sentence. When we reach the end of the s, add sentence to sentences. Then we need to recover sentence back to its original status before word was added. So I simply record a temporary string temp each time before I add word to sentence and use it to recover the status.

The code is as follows. I hope it to be self-explanatory enough.

 1 class Solution {
 2 public:
 3     bool isWordBreak(string& s, unordered_set<string>& wordDict) {
 4         vector<bool> dp(s.length() + 1, false);
 5         dp[0] = true;
 6         int minlen = INT_MAX;
 7         int maxlen = INT_MIN;
 8         for (string word : wordDict) {
 9             minlen = min(minlen, (int)word.length());
10             maxlen = max(maxlen, (int)word.length());
11         }
12         for (int i = 1; i <= s.length(); i++) {
13             for (int j = i - minlen; j >= max(0, i - maxlen); j--) {
14                 if (dp[j] && wordDict.find(s.substr(j, i - j)) != wordDict.end()) {
15                     dp[i] = true;
16                     break;
17                 }
18             }
19         }
20         return dp[s.length()];
21     }
22     void breakWords(string s, int idx, unordered_set<string>& wordDict, string& sol, vector<string>& res) {
23         if (idx == s.length()) {
24             sol.resize(sol.length() - 1);
25             res.push_back(sol);
26             return;
27         }
28         for (int i = idx; i < s.length(); i++) {
29             string word = s.substr(idx, i - idx + 1);
30             if (wordDict.find(word) != wordDict.end()) {
31                 string tmp = sol;
32                 sol += word + " ";
33                 breakWords(s, i + 1, wordDict, sol, res);
34                 sol = tmp;
35             }
36         }
37     }
38     vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
39         string sol;
40         vector<string> res;
41         if (!isWordBreak(s, wordDict)) return res;
42         breakWords(s, 0, wordDict, sol, res);
43         return res;
44     }
45 };