139. Word Break 动态规划

Given a string s and a dictionary of words dict, determine ifs can be segmented into a space-separated sequence of one or more dictionary words.

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

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

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分析：

通过动态规划进行求解，以“leetcode"求解,以s为对象进行处理，若结果为true,则必存在一个开头在list中，因此假设存在这样一个在list中在s的j位置，因此下一个应该找到从j向后走的，重复前面一个动作。一直到s的最后，若能找到则dp[s.size()]=true,否则为false.

代码：

class Solution {
public:
    bool wordBreak(string s, unordered_set& word) {
     if(word.size()==0) return false;
     vectordp (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 temp=s.substr(j,i-j);
             if(word.find(temp)!=word.end())
             {
                 dp[i]=true;
                 break;
             }
         }
     }
     return dp[s.size()];
    }
};