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

原文：https://www.cnblogs.com/ruruozhenhao/p/10367558.html