LEETCODE: Word Break II

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"].

创建一张二维表记录 s(i,j) 是否是dict里的一个元素。后面大家就知道了。参考：http://blog.youkuaiyun.com/lanxu_yy/article/details/17310247

class Solution {
public:
    vector<vector<bool>> dptable;
    vector<string> results;
    vector<string> records;
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        dptable = vector<vector<bool>>(s.length(),vector<bool>());
        for(int ii = 0; ii < s.length(); ii ++) {
            for(int jj = ii; jj < s.length(); jj ++) {
                dptable[ii].push_back(isMatch(s.substr(ii, jj - ii + 1), dict)); 
            }
        }
        findresults(s.length() - 1, s);
        return results;
    }
    void findresults(int ii, string s) {
        if(ii == -1) {
            string str;
            for(int jj = records.size() - 1; jj >= 0; jj --) {
                str += records[jj];
                if(jj != 0) {
                    str.push_back(' ');
                }
            }
            results.push_back(str);
        }
        else {
            for(int kk = 0; kk <= ii; kk ++) {
                if(dptable[kk][ii - kk]) {
                    records.push_back(s.substr(kk, ii - kk + 1));
                    findresults(kk - 1, s);
                    records.pop_back();
                }
            }
        }
    }
    bool isMatch(string str, unordered_set<string> &dict) {
        unordered_set<string>::const_iterator got = dict.find(str);
        if(got != dict.end()) {
            return true;
        }
        else {
            return false;
        }
    }
};