题目：

Given a string s and a dictionary of words dict, determine if s 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".

思路：

用NP的思路可以做，不过时间复杂度比较高。因此可以考虑使用动态规划DP的方式做。

代码：

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

或者简洁一点

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