[LeetCode]139.Word Break

题目

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

递归

代码

/*---------------------------------------*   日期：2015-05-07*   作者：SJF0115*   题目: 139.Word Break*   网址：https://leetcode.com/problems/word-break/*   结果：AC*   来源：LeetCode*   博客：-----------------------------------------*/#include <iostream>#include <vector>#include <unordered_set>using namespace std;class Solution {public:    bool wordBreak(string s, unordered_set<string>& wordDict) {        int size = wordDict.size();        if(size == 0){            return false;        }//if        vector<bool> visited(s.size(),false);        return helper(s,wordDict,0,visited);    }private:    bool helper(string &s,unordered_set<string>& wordDict,int index,vector<bool> &visited){        if(index >= s.size()){            return true;        }//if        // 已经查看过表示行不通        if(visited[index]){            return false;        }//if        visited[index] = true;        // 以index下标开始        for(int i = index;i < s.size();++i){            if(wordDict.find(s.substr(index,i - index + 1)) != wordDict.end()){                if(helper(s,wordDict,i+1,visited)){                    return true;                }//if            }//if        }//for        return false;    }};int main() {    Solution solution;    string str("leetcode");    unordered_set<string> wordDict = {"leet","co","code"};    cout<<solution.wordBreak(str,wordDict)<<endl;}

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