LeetCode--Wrod Break

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class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        int nsize=s.size();
        int i=0,j=0;
        bool *dp = new bool[nsize];
        memset(dp,false,sizeof(dp));
        
        for(i=0;i<nsize;++i)
        {
            dp[i] = ((dict.find(s.substr(0,i+1))!=dict.end())?true:false);
            if(dp[i])
                continue;
            else
            {
                for(j=0;j<i;++j)
                {
                    if(dp[j])
                    {
                        dp[i] = ((dict.find(s.substr(j+1,i-j))!=dict.end())?true:false) | dp[i];
                    }
                }
            }
        }
        return dp[nsize-1];
        delete []dp;
    }
};

递归到Dp转化的经典题目

递归如下

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        if(s.length() == 0)
        {
            return false;
        }
        for(int i = 1 ; i <= s.length() ; ++i)
        {
            string word = s.substr(0,i);
            string str = s.substr(i,s.length()-i);
            if(dict.find(word) != dict.end() && wordBreak(str,dict))
            {
                return true;
            }
        }
        return false;
    }
};

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