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题目如下：
给定一个字符串s和一组单词dict，判断s是否可以用空格分割成一个单词序列，使得单词序列中所有的单词都是dict中的单词（序列可以包含一个或多个单词）。
例如:
给定s=“leetcode”；
dict=["leet", "code"].
返回true，因为"leetcode"可以被分割成"leet code".
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".
此题可以用动态规划来做，因为不知道这个单词到底有多少个词组成，所以我们把这个问题化为n个小问题，然后以小问题逐个推导到后面的答案。
class Solution { public: bool wordBreak(string s, unordered_set<string> &dict)
{ if (s.empty()){ return false; } if (dict.empty()){ return false; }
vector<bool>a(s.size()+1, 0); a[0] = true; for (int i = 1; i<s.size()+1; i++){
for (int j = 0; j<i; j++){ if (a[j]){ if (dict.count(s.substr(j, i - j))){ a[i]
= true; break; } } } } return a[s.size()]; } };
