leetcode || 140、Word Break II

problem：

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"].

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 Dynamic Programming Backtracking

thinking:

1 使用二维表vector<vector<int> >tbl记录，记录从i点，能否跳到下一个break位置。如果不能，那么tbl[i]就为空。如果可以，就记录可以跳到哪些位置。
2 利用二维表优化递归回溯法。
优化点：
如果当前位置是start，但是tbl[start]为空，那么就是说，这个break位置不能break整个s串的，直接返回上一层，不用搜索到下一层了。

code:

class Solution {public:vector<string> wordBreak(string s, unordered_set<string> &dict) {vector<string> rs;string tmp;vector<vector<int> > tbl = genTable(s, dict);word(rs, tmp, s, tbl, dict);return rs;}void word(vector<string> &rs, string &tmp, string &s, vector<vector<int> > &tbl,unordered_set<string> &dict, int start=0){if (start == s.length()){rs.push_back(tmp);return;}for (int i = 0; i < tbl[start].size(); i++){string t = s.substr(start, tbl[start][i]-start+1);if (!tmp.empty()) tmp.push_back(' ');tmp.append(t);word(rs, tmp, s, tbl, dict, tbl[start][i]+1);while (!tmp.empty() && tmp.back() != ' ') tmp.pop_back();//tmp.empty()if (!tmp.empty()) tmp.pop_back();}}vector<vector<int> > genTable(string &s, unordered_set<string> &dict){int n = s.length();vector<vector<int> > tbl(n);for (int i = n - 1; i >= 0; i--){if(dict.count(s.substr(i))) tbl[i].push_back(n-1);}for (int i = n - 2; i >= 0; i--){if (!tbl[i+1].empty())//if we can break i->n{for (int j = i, d = 1; j >= 0 ; j--, d++){if (dict.count(s.substr(j, d))) tbl[j].push_back(i);}}}return tbl;}};