[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];
    }
};

　　

posted @ 2014-01-07 15:14  七年之后  阅读(216)  评论(0)    收藏  举报