17. Word Break && Word Break II

Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given s = "leetcode", dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

说明： 深度搜索，一定要记忆下每次走完的结果（此处记下筛掉的情况）。

bool judge(string s, unordered_set<string> &dict, vector<bool> &tag) {
    if(s == "") return true;
    for(int i = 1; i <= s.length(); ++i) {
        if(tag[s.size()-i] && dict.find(s.substr(0, i)) != dict.end())  {
            if (judge(s.substr(i, s.size()-i), dict, tag)) return true;
            else tag[s.size()-i] = 0;
        }
    }
    return false;
}

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        if(s == "") return true;
        vector<bool> tag(s.size()+1, true); //the value is the result that (index) length of reserved string can return;
        return judge(s, dict, tag);
    }
};

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"].

说明： 方法比较巧妙。记忆下每个位置开始的所有能成回文串的结束位置。然后深搜。

void dfs(string s, vector<vector<int> > & Reach, int Id, string path, vector<string> &vec) {
	if(Id == s.size()) { vec.push_back(path); return; }
	for(size_t i = 0; i < Reach[Id].size(); ++i) {
		path = path + (Id == 0 ? s.substr(Id, Reach[Id][i]) : " " + s.substr(Id, Reach[Id][i]-Id));
		dfs(s, Reach, Reach[Id][i], path, vec);
		path.erase(path.end()-(Id == 0 ? Reach[Id][i] : (Reach[Id][i]-Id+1)), path.end());
	}
}
class Solution {
public:
	vector<string> wordBreak(string s, unordered_set<string> &dict) {
		vector<string> vec;
		int n = s.size();
		if(n == 0) return vec;
		vector<vector<int> > reachable(n, vector<int>());
		for(int end = n; end > 0; --end) {
			if(end < n && reachable[end].empty()) continue;
			for(int start = 0; start < end; ++start) {
				if(dict.find(s.substr(start, end-start)) != dict.end())
					reachable[start].push_back(end);
			}
		}
		dfs(s, reachable, 0, string(""), vec);
		return vec;
	}
};

两题公有的易超时反例：aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaa……aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab