标签：style http color os io ar for art div
Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
Return all such possible sentences.
For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].
A solution is ["cats and dog",
"cat sand dog"].
题意：
给定一个字符串和一个单词字典，找出可以用字典里的单词来分割这个字符串的各种组合。
思路：dfs
void dfs(int start, string partition);
start表示当前要分割的字符串s[start:]，
设 k将 s分割为s[:k]和s[k:]，则 0 < k <= s.size() 且 s[:k]存在字典dict中
按上面这样的思路直接做会超时，可采用dfs + 剪枝的思路
增加一个bool数组 possible[k] 表示字符串s[k:]是否可用字典里的单词来分割
unordered_set<string> _dict;
vector<string> res;
vector<bool> possible;
string _s;
void dfs(int start, string partition){
if(start >= _s.size()){
res.push_back(partition.substr(1, partition.size() - 1));
return;
}
for(int k = start + 1; k <= _s.size(); ++k){
if(possible[k] && _dict.find(_s.substr(start, k - start)) != _dict.end()){ //剪枝
int size_before = res.size();
dfs(k , partition + " " + _s.substr(start, k - start));
if(res.size() == size_before) possible[k] = false; //剪枝
}
}
}
vector<string> wordBreak(string s, unordered_set<string> &dict){
_dict = dict;
_s = s;
possible = vector<bool>(s.size() + 1, true);
dfs(0, "");
return res;
}标签：style http color os io ar for art div
原文地址：http://blog.csdn.net/zhengsenlie/article/details/39164127
