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.

这题比较有意思，并且在NLP处理中可能会应用很多，所以这道题可以深入挖掘。比如改进字符串匹配程序，可以加快它的查找速率。

如果dp[j-1] = true；（即stirng[0,…,j-1]匹配成功），如果子串string[j,..,i]能够匹配成功，则dp[i] = true,即整个字符串能够匹配成功。

class Solution {
public:
bool searchWork(string s,vector<string>& wordDict) {
for(int i = 0;i < wordDict.size(); ++i){
if(s == wordDict[i]){
return true;
}
}

return fals
4000
e;
}

bool wordBreak(string s, vector<string>& wordDict) {
int n = s.size();
vector<bool> dp(n+1,false);

if(n <= 0){
return true;
}

dp[0] = true;
for(int i = 1;i <= n;++i){
for(int j = 1;j <= i;++j){
if(searchWork(s.substr(j-1,i-j+1),wordDict)){
dp[i] = dp[i] || dp[j-1];*强调内容*
}
}
}

return dp
;
}
};