dict = ["cat", "cats", "and", "sand", "dog"]. A solution is ["cats and dog", "cat sand dog"].*/classSolution {public: vector<string> wordBreak(strings, unordered_set<string> &dict) { vector<string>retvec;if(s.size() ==0|| dict.size() ==0)returnretvec;intlen =s.size...
Can you solve this real interview question? Word Break II - Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order. Note
}//here we got//1 . whether string breakable wordbreak[n],//2. lookup matrix like a graphif(!wordbreak[size])returnresult; backtrack(s, lookup, size, result, path);returnresult; }
Can you solve this real interview question? Word Break II - Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order. Note
A solution is["cats and dog", "cat sand dog"]. 难度是5星级了。 难度指数直逼word ladder II了 主要的难点: 要优化到极致,不能直接使用递归回溯法,否则超时。 本程序方法: 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...
解题思路同 LeetCode: 139. Word Break 题解。 此题不同之处在于,需要将计算过程记录下来,后面根据记录的计算过程构造最优解。 AC 代码 class Solution { private: // Compute the value of an optimal solution, in a bottom-up fashion. ...
Return true because"leetcode"can be segmented as"leet code". 博文Word Break -- LeetCode提及了动态规划的思想: 首先我们要决定要存储什么历史信息以及用什么数据结构来存储信息。然后是最重要的递推式,就是如从存储的历史信息中得到当前步的结果。最后我们需要考虑的就是起始条件的值。
代码: importjava.util.ArrayList;importjava.util.Set;publicclassSolution{privateArrayList<String>resultList;privateArrayList<String>tmpList;publicArrayList<String>wordBreak(Strings,Set<String>dict){resultList=newArrayList();tmpList=newArrayList();divide(s,dict);returnresultList;}privatevoiddivide(Strings,Set...
参考了Discuss区一个DFS+DP的解法(https://leetcode.com/problems/word-break-ii/discuss/44262/My-C%2B%2B-DP%2BDFS-solution-(4ms)),学到了一种新的剪枝策略: classSolution{private://DFS path build functionvoidbuildPath(boolisBreakable[],string&s,intpos,vector<string>&res,string curP,unordered_...
[LeetCode] 126. Word Ladder II Problem Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary...