【Leetcode】139. Word Break（字符串分块）

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Given a non-empty string s and a dictionary wordDictcontaining 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

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true

Example 3:

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

题目大意：

给出一个字符串和一个字典，要求从字典当中查询出word来组合成目标s。

解题思路：

我们看到下面的两个函数，将字符串分成两个部分，递归判断这两部分在字典中是否有对应word。

str.substr（a）--截取str里面从a开始之后的字符组成的字符串

str.substr（a，b） --截取str里面从a开始之后的字符长度为b的字符串

class Solution {
private:
    unordered_map<string, int> m;
    unordered_map<string, bool> dp;
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        for(auto x: wordDict) m[x]++;
        return DFS(s);
    }
    
    bool DFS(string s){
        if(dp.count(s)) return dp[s];
        if(s.empty()) return true;
        bool found=false;
        for(int i=1;i<=s.size() && !found;i++){
            if(m.count(s.substr(0,i))) found |= DFS(s.substr(i));
        }
        dp[s] = found;
        return dp[s];
    }
    
};