标签：
I
title:
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".
思路：一开始就考虑直接使用dfs超时了。。。
class Solution { public: bool wordBreak(string s, unordered_set<string>& wordDict) { if (s.size() == 0) return true; string tmp; for (int i = 0; i < s.size(); i++){ tmp.push_back(s[i]); if (wordDict.find(tmp) != wordDict.end()){ string tmp1(s.begin()+i+1,s.end()); if (wordBreak(tmp1,wordDict)) return true; } } return false; } };
其实对于这样的问题，是可以想到是可以使用动态规划的。
一个DP问题。定义possible[i] 为S字符串上[0,i]的子串是否可以被segmented by dictionary.
那么
possible[i] = true if S[0,i]在dictionary里面
= true if possible[k] == true 并且 S[k+1,j]在dictionary里面， 0<k<i
= false if no such k exist.
class Solution { public: bool wordBreak(string s, unordered_set<string>& wordDict) { int len = s.size(); vector<bool> dp(len+1,false); dp[0] = true; for (int i = 1; i < len+1; i++){ for (int k = 0; k < i; k++){ if (dp[k] && wordDict.find(s.substr(k,i-k)) != wordDict.end()){ dp[i] = true; break;//break是因为题目只是需要判断是否存在 } } } return dp[len]; } };
II
title:
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"].
思路：使用dp，并且保存中间结果。注意，在I中找到一个解就直接break，这里需要记录下这个信息。另外prev[i][j]为true表示字串[j,i]在字典中
class Solution { public: vector<string> wordBreak(string s, unordered_set<string>& wordDict) { int len = s.size(); vector<bool> dp(len+1,false); vector<vector<bool> > prev(len+1,vector<bool>(len,false)); dp[0] = true; for (int i = 1; i < len+1; i++){ for (int k = 0; k < i; k++){ if (dp[k] && wordDict.find(s.substr(k,i-k)) != wordDict.end()){ dp[i] = true; prev[i][k] = true; } } } vector<string> result; vector<string> results; genPath(len,s,dp,prev,result,results); return results; } void genPath(int cur,string s, vector<bool> dp,vector<vector<bool> > prev, vector<string> &result,vector<string> &results){ if (cur == 0){ string tmp; for (int i = result.size()-1; i > 0; i--){ tmp += result[i] + " "; } tmp += result[0]; results.push_back(tmp); return ; } for (int i = 0; i < s.size(); i++){ if (prev[cur][i]){ result.push_back(s.substr(i,cur-i)); genPath(i,s,dp,prev,result,results); result.pop_back(); } } } };
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原文地址：http://www.cnblogs.com/yxzfscg/p/4525074.html
