[LeetCode]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".

思路：回溯+递归=超时

 1 class Solution {

 2 public:

 3     bool wordBreak(string s, unordered_set<string> &dict) {

 4         if(dict.find(s)!=dict.end()) return true;

 5         int len=s.size();

 6         for(int i=1;i<=len;i++)

 7         {

 8             if(dict.find(s.substr(0,i))!=dict.end())

 9             {

10                 string str=s.substr(i,len-i);

11                 //return wordBreak(str,dict);

12                 if(wordBreak(str,dict)) return true;

13             }

14         }

15         return false;

16     }

17 };

View Code

备忘录法，记忆化搜索。

class Solution {

public:

    bool wordBreak(string s, unordered_set<string> &dict) {

        if(dict.find(s)!=dict.end()) return true;

        int len=s.size();

        vector<bool> dp(len+1,false);

        dp[0]=true;

        for(int i=1;i<=len;i++)

        {

            for(int j=0;j<i;j++)

            {

                if(dp[i]=dp[j]&&(dict.find(s.substr(j,i-j))!=dict.end()))

                    break;

            }

        }

        return dp[len];

    }

};