[LeetCode] WordBreak II, Solution

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"]

. [Thoughts] 既然是要给出所有解，那就递归好了，但是递归的过程中注意剪枝。

  没有剪枝版

[code]     vector<string> wordBreak(string s, unordered_set<string> &dict) {
string result;
vector<string> solutions;
int len = s.size();
GetAllSolution(0, s, dict, len, result, solutions);
return solutions;
}

void GetAllSolution(int start, const string& s, const unordered_set<string> &dict, int len, string& result, vector<string>& solutions)
{
if (start == len)
{
solutions.push_back(result.substr(0, result.size()-1));//eliminate the last space
return;
}
for (int i = start; i< len; ++i)
{
string piece = s.substr(start, i - start + 1);
if (dict.find(piece) != dict.end() && possible[i+1])
{
result.append(piece).append(" ");
GetAllSolution(i + 1, s, dict, len, result, solutions);
result.resize(result.size() - piece.size()-1);
}
}
}

但是很明显，这种裸的递归，效率太低，因为有大量的重复计算，肯定过不了测试数据。
剪枝版
这里加上一个possible数组，如同WordBreak I里面的DP数组一样，用于记录区间拆分的可能性

Possible[i] = true 意味着 [i,n]这个区间上有解

加上剪枝以后，运行时间就大大减少了。（Line 5, 20, 23, 25,26为剪枝部分）

vector<string> wordBreak(string s, unordered_set<string> &dict) {
string result;
vector<string> solutions;
int len = s.size();
vector<bool> possible(len+1, true); // to record the search which has been executed once
GetAllSolution(0, s, dict, len, result, solutions, possible);
return solutions;
}

void GetAllSolution(int start, const string& s, const unordered_set<string> &dict, int len, string& result, vector<string>& solutions, vector<bool>& possible)
{
if (start == len)
{
solutions.push_back(result.substr(0, result.size()-1));//eliminate the last space
return;
}
for (int i = start; i< len; ++i)
{
string piece = s.substr(start, i - start + 1);
if (dict.find(piece) != dict.end() && possible[i+1]) // eliminate unnecessory search
{
result.append(piece).append(" ");
int beforeChange = solutions.size();
GetAllSolution(i + 1, s, dict, len, result, solutions, possible);
if(solutions.size() == beforeChange) // if no solution, set the possible to false
possible[i+1] = false;
result.resize(result.size() - piece.size()-1);
}
}
}

suoyi