139. Word Break

问题：

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"

.

分析：熊签到后依次遍历，看是否当前字母所在的单词是属于dict的，若不属于时，退出循环，返回false，若循环结束，则返回true

代码：

class Solution {

public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
int len=s.size();
vector<bool> dp(len+1,false);
dp[0]=true;
for(int i=1;i<=len;i++){
for(int j=i-1;j>=0;j--){
if(dp[j]&&wordDict.count(s.substr(j,i-j))){
dp[i]=true;
break;
}
}
}
return dp[len];
}

};