[leetcode-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.
You may assume the dictionary does not contain duplicate words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

参考思路：

bool wordInDict(string s,vector<string>& wordDict)
    {
        if (s == "")return true;
        vector<string>::iterator it;
        for (it = wordDict.begin(); it != wordDict.end();it++)
            if (*it == s)    return true;        
     return false;
    }
    bool wordBreak(string s, vector<string>& wordDict)
    {
        vector<bool> flag(s.size(),false);//从某位置开始，是否能分词
        return wordBreak(s, wordDict, 0, flag);
    }
    bool wordBreak(string s, vector<string>& wordDict,int index,vector<bool>& flag)
    {
        if (flag[index]) return false;
        
        int len = s.size() - index;
        for (int i = len; i > 0;i--)
        {
            string word = s.substr(index, i);
            if (wordInDict(word,wordDict))
            {
                if (i == len) return true;
                elseif (wordBreak(s, wordDict, index+i, flag)) return true;                
            }
        }
        flag[index] = true;
        return false;
    }

 学习了LeetCode上牛人写的动态规划版本。

bool wordBreak(string s, unordered_set<string> &dict) {
        if(dict.size()==0) return false;
        
        vector<bool> dp(s.size()+1,false);
        dp[0]=true;
        
        for(int i=1;i<=s.size();i++)
        {
            for(int j=i-1;j>=0;j--)
            {
                if(dp[j])
                {
                    string word = s.substr(j,i-j);
                    if(dict.find(word)!= dict.end())
                    {
                        dp[i]=true;
                        break; //next i
                    }
                }
            }
        }
        
        return dp[s.size()];
    }

 

 

参考：

http://www.jianshu.com/p/f30581e8d343

https://discuss.leetcode.com/topic/7299/c-dynamic-programming-simple-and-fast-solution-4ms-with-optimization