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"

.

UPDATE (2017/1/4):

The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> dict;
for(int i=0;i<wordDict.size();i++){
dict.insert(wordDict[i]);
}
return towordBreak(s,dict);
}
bool towordBreak(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()];
}

};

我们使用一个布尔向量dp []。 如果有效字（单词序列）结束，那么dp [i]设置为true。 优化是从当前位置我回来，只有子字符串和字典查找，以防发现前面的位置j与dp [j] == true。