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.
一般与字符串有关的题都要运用动态规划，此题中dp[i]为true代表从0到i-1位可以成功分割，所以递推式为dp[i]=dp[j]&&(dict中存在j+1到i位构成的子串)。
注意此处int n=s.size();不要写成int n=wordDict.size();debug好久，又是低级错误。

class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
if(s.empty())
return true;
int n=s.size();
vector<bool> dp(n+1,false);
dp[0]=true;
for(int i=1;i<=n;i++)
{
for(int j=0;j<i;j++)
{
if(dp[j]&&find(wordDict.begin(),wordDict.end(),s.substr(j,i-j))!=wordDict.end())
{
cout<<i;
dp[i]=true;
break;
}
}
}
return dp
;
}
};