139. Word Break

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本文探讨了WordBreak问题的解决方法，通过使用动态规划（DP）算法，有效地判断一个字符串是否能被拆分为字典中的一系列单词。示例包括leetcode和applepenapple的成功案例，以及catsandog的失败案例。

139. Word Break

Medium

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Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

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这道题可以用记忆化和dp来做，这里用了dp算法，一开始要用hashset保存一下，dp[i]表示[0,i)字符串是否满足可以在hashset中找到对应的分割单词

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
        vector<bool> dp(s.size() + 1);
        dp[0] = true;
        for (int i = 0; i < dp.size(); ++i) {
            for (int j = 0; j < i; ++j) {
                if (dp[j] && wordSet.count(s.substr(j, i - j))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp.back();
    }
};