[Leetcode] 139. Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

　　题目大意：给定字符串s与字典dict，判断字符串s能否由字典中的元素组成。

　　分析：使用动态规划法。dp[i]=true表示字符串中0到i-1部分的子字符串能由字典中字符串组成。

 

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int n=s.size();
        vector<bool> dp(n+1, false);
        dp[0]=true;
        for(int i=1;i<=n;i++)
        {
            for(int j=i-1;j>=0;j--)
            {
                if(dp[j])
                {
                    string sub = s.substr(j,i-j);
                    if(findWord(sub, wordDict))
                    {
                        dp[i]=true;
                        break;
                    }     
                }
            }
            
        }
        return dp[n];
    }
    bool findWord(string s, vector<string>& wordDict)
    {
        for(int i=0;i<wordDict.size();i++)
        {
            if(s==wordDict[i]) return true;
        }
        return false;
    }
};