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.

我的解法是DFS，但在Discuss中还有dp的解法，一开始用的递归但是没写出来，后面想了很久还是用while+队列的形式解的

此题的DFS思路就是  先用start 与 end进行切割，如找到了leet，则在leet的基础上令start=end+1,end继续往后面循环，直到end超过string.size()-1

小细节是需要用增加一个记录start的无序表，避免超时

class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
if(wordDict.size()==0||s.size()==0)
return false;
unordered_set<int> visited;    //一开始没这个东西，但是超时了
queue<int> q;
q.push(0);
while(!q.empty()){
int start=q.front();
q.pop();
if(visited.find(start)==visited.end()){
visited.insert(start);
for(int end=start;end<s.size();end++){
string temp=s.substr(start,end-start+1);
if(find(wordDict.begin(),wordDict.end(),temp)!=wordDict.end()){
q.push(end+1);
if(end+1==s.size())
return true;
}
}
}
}
return false;
}
};

DP版本的：

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

python版本：

class Solution:
def wordBreak(self, s, wordDict):
"""
:type s: str
:type wordDict: List[str]
:rtype: bool
"""
dp=[False]*(len(s)+1)
dp[0]=True
for i in range(1,len(s)+1):
for j in range(i):
if dp[j] and s[j:i] in wordDict:
dp[i]=True
return dp[len(s)]