Word Break



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".

常规思维用个递归就可以了，但这里有时间限制，最好别用递归。最好的方法是使用动态规划，

class Solution {
public:
	bool wordBreak(string s, unordered_set<string> &dict) {
		if(!s.size()) return false;
		vector<int> satisfied(s.size(),0);//记录s[0-i]是否可以分词
		for(size_t i=0;i<s.size();++i){
			string ts = s.substr(0,i+1);
			unordered_set<string>::iterator iter= dict.find(ts);
			if(iter!=dict.end()){
				satisfied[i]=1;
			}else{
				for(size_t j=0;j<i;++j){
					if(!satisfied[j]) continue;
					iter = dict.find(s.substr(j+1,i-j));
					if(iter!=dict.end()){
						satisfied[i]=1;
						break;
					}
				}
			}
		}

		return satisfied[s.size()-1];
	}
};