[LeetCode] Word Break, Solution

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, given
s =

"leetcode"

,
dict =

["leet", "code"]

. Return true because

"leetcode"

can be segmented as

"leet code"

. [Thoughts] 一个DP问题。定义possible[i] 为S字符串上[0,i]的子串是否可以被segmented by dictionary. 那么 possible[i] = true if S[0,i]在dictionary里面 = true if possible[k] == true 并且 S[k+1,j]在dictionary里面， 0<k<i = false if no such k exist.

 实现时前面加一个dummy节点，这样可以把三种情况统一到一个表达式里面。

[code]     bool wordBreak(string s, unordered_set<string> &dict) {
string s2 = '#' + s;
int len = s2.size();
vector<bool> possible(len, 0);

possible[0] = true;
for(int i =1; i< len; ++i)
{
for(int k=0; k<i; ++k)
{
possible[i] = possible[k] &&
dict.find(s2.substr(k+1, i-k)) != dict.end();
if(possible[i]) break;
}
}

return possible[len-1];
}