Word Break

  • tags: [DP_Sequence]

Question

  1. Given a string s and a dictionary of words dict, determine if s can be
  2. segmented into a space-separated sequence of one or more dictionary words.
  4. For example, given
  5. s = "leetcode",
  6. dict = ["leet", "code"].
  8. Return true because "leetcode" can be segmented as "leet code".

题解

单序列(DP_Sequence) DP 题，由单序列动态规划的四要素可大致写出：

  1. State: f[i] 表示前i个字符能否根据词典中的词被成功分词。
  2. Function: f[i] = or{f[j], j < i, letter in [j+1, i] can be found in dict}, 含义为小于i的索引j中只要有一个f[j]为真且j+1到i中组成的字符能在词典中找到时，f[i]即为真，否则为假。具体实现可分为自顶向下或者自底向上。
  3. Initialization: f[0] = true, 数组长度为字符串长度 + 1，便于处理。
  4. Answer: f[s.length]

考虑到单词长度通常不会太长，故在s较长时使用自底向上效率更高。

Python

  1. class Solution:
  2.     # @param s, a string
  3.     # @param wordDict, a set<string>
  4.     # @return a boolean
  5.     def wordBreak(self, s, wordDict):
  6.         if not s:
  7.             return True
  8.         if not wordDict:
  9.             return False
  11.         max_word_len = max([len(w) for w in wordDict])
  12.         can_break = [True]
  13.         for i in xrange(len(s)):
  14.             can_break.append(False)
  15.             for j in xrange(i, -1, -1):
  16.                 # optimize for too long interval
  17.                 if i - j + 1 > max_word_len:
  18.                     break
  19.                 if can_break[j] and s[j:i + 1] in wordDict:
  20.                     can_break[i + 1] = True
  21.                     break
  22.         return can_break[-1]

C++

  1. class Solution {
  2. public:
  3.     bool wordBreak(string s, unordered_set<string>& wordDict) {
  4.         if (s.empty()) return true;
  5.         if (wordDict.empty()) return false;
  7.         // get the max word length of wordDict
  8.         int max_word_len = 0;
  9.         for (unordered_set<string>::iterator it = wordDict.begin();
  10.          it != wordDict.end(); ++it) {
  12.             max_word_len = max(max_word_len, (*it).size());
  13.         }
  15.         vector<bool> can_break(s.size() + 1, false);
  16.         can_break[0] = true;
  17.         for (int i = 1; i <= s.size(); ++i) {
  18.             for (int j = i - 1; j >= 0; --j) {
  19.                 // optimize for too long interval
  20.                 if (i - j > max_word_len) break;
  22.                 if (can_break[j] && 
  23.             wordDict.find(s.substr(j, i - j)) != wordDict.end()) {
  25.                     can_break[i] = true;
  26.                     break;
  27.                 }
  28.             }
  29.         }
  31.         return can_break[s.size()];
  32.     }
  33. };

Java

  1. public class Solution {
  2.     public boolean wordBreak(String s, Set<String> wordDict) {
  3.         if (s == null || s.length() == 0) return true;
  4.         if (wordDict == null || wordDict.isEmpty()) return false;
  6.         // get the max word length of wordDict
  7.         int max_word_len = 0;
  8.         for (String word : wordDict) {
  9.             max_word_len = Math.max(max_word_len, word.length());
  10.         }
  12.         boolean[] can_break = new boolean[s.length() + 1];
  13.         can_break[0] = true;
  14.         for (int i = 1; i <= s.length(); i++) {
  15.             for (int j = i - 1; j >= 0; j--) {
  16.                 // optimize for too long interval
  17.                 if (i - j > max_word_len) break;
  19.                 String word = s.substring(j, i);
  20.                 if (can_break[j] && wordDict.contains(word)) {
  21.                     can_break[i] = true;
  22.                     break;
  23.                 }
  24.             }
  25.         }
  27.         return can_break[s.length()];
  28.     }
  29. }

源码分析

Python 之类的动态语言无需初始化指定大小的数组，使用时下标i比 C++和 Java 版的程序少1。使用自底向上的方法求解状态转移，首先遍历一次词典求得单词最大长度以便后续优化。

复杂度分析

  1. 求解词典中最大单词长度，时间复杂度为词典长度乘上最大单词长度 $$O(L_D \cdot L_w)$$
  2. 词典中找单词的时间复杂度为 $$O(1)$$(哈希表结构)
  3. 两重 for 循环，内循环在超出最大单词长度时退出，故最坏情况下两重 for 循环的时间复杂度为 $$O(n L_w)$$.
  4. 故总的时间复杂度近似为 $$O(n L_w)$$.
  5. 使用了与字符串长度几乎等长的布尔数组和临时单词word，空间复杂度近似为 $$O(n)$$.