# 139. Word Break

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```

Approach #1:

```class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> _wordDict(wordDict.begin(), wordDict.end());
return wordBreak(s, _wordDict);
}

bool wordBreak(const string& s, const unordered_set<string>& wordDict) {
if (memo_.count(s)) return memo_[s];
if (wordDict.count(s)) return memo_[s] = true;
for (int j = 1; j < s.length(); ++j) {
string left = s.substr(0, j);
string right = s.substr(j);
if (wordDict.count(right) && wordBreak(left, wordDict))
return memo_[s] = true;
}
return memo_[s] = false;
}

private:
unordered_map<string, bool> memo_;
};
```

139. Word Break

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