140. Word Break II
题目:
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.
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"].
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.
解析
//运行时间：4ms
//占用内存：508k
class Solution {
vector<string> combine(string word, vector<string> prev){
for (int i = 0; i < prev.size(); ++i){
prev[i] += " " + word;
}
return prev;
}
public:
vector<string> wordBreak(string s, unordered_set<string>& dict) {
vector<string> result;
if (dict.count(s)){ //a whole string is a word
result.push_back(s);
}
for (int i = 1; i < s.size(); ++i){
string word = s.substr(i);
if (dict.count(word)){
string rem = s.substr(0, i);
vector<string> prev = combine(word, wordBreak(rem, dict));
result.insert(result.end(), prev.begin(), prev.end());
}
}
reverse(result.begin(), result.end());
return result;
}
};
namespace test
{
vector<string> wordBreak(string s, unordered_set<string> &dict) {
//暴力搜索，不能ac，复杂度超了。
vector<string> res;
int size = s.length();
for (int i = 0; i < size; i++){
string tmp = s.substr(0, i + 1);
if (dict.count(tmp))
findNext(res, s, dict, tmp, i + 1);
}
return res;
}
void findNext(vector<string> & res, string s, unordered_set<string> &dict, string tmp, int i){
int size = s.length();
if (i >= size){
res.push_back(tmp);
return;
}
for (int j = 1; j <= size - i; ++j){
string now = s.substr(i, j);
if (dict.count(now)){
findNext(res, s, dict, tmp + ' ' + now, i + j);
}
}
}
}
题目来源