139. Word Break

s and a dictionary of words dict, determine if s

For example, given

s = ​​"leetcode"​​,

dict = ​​["leet", "code"]​​.​​"leetcode"​​ can be segmented as ​​"leet code"​​.

设dp[i]为前i个字符是否可以切割。

则dp[i]=dp[j]&&s.substr(j,i-j)

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


140. Word Break II


s and a dictionary of words dict, add spaces in s

Return all such possible sentences.

For example, given

s = ​​"catsanddog"​​,

dict = ​​["cat", "cats", "and", "sand", "dog"]​​.​​["cats and dog", "cat sand dog"]​​.



用记忆递归


思路：不断切两刀，后面的有效，再递归前面的。 




class Solution {
public:
  vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
    if(m.count(s)){
      return m[s];
    }
    vector<string> res;
    if(wordDict.count(s)){//此时s直接在字典中
      res.push_back(s);
    }
    for(int i=1;i<s.size();i++){
      string last=s.substr(i);
      if(wordDict.count(last)){
        string pre=s.substr(0,i);
        vector<string> tmp=combine(last,wordBreak(pre,wordDict));
        res.insert(res.end(),tmp.begin(),tmp.end());
      }
    }
    m[s]=res;//记忆
    return res;
  }
private:
  vector<string> combine(string last,vector<string> pre){
    for(int i=0;i<pre.size();i++){
      pre[i]+=" "+last;
    }
    return pre;
  }

  unordered_map<string,vector<string>> m;
};