139. Word Break

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139. Word Break

• 题目

 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. 

• 解析

class Solution {
public:
	bool wordBreak(string s, vector<string>& wordDict) {

		set<string> strset(wordDict.begin(), wordDict.end());//vector-->set

		if (wordDict.size() == 0)
		{
			return false;
		}
		//动态规划
		vector<bool> dp(s.size() + 1, false);
		dp[0] = true;

		for (int i = 1; i <= s.size(); i++) //第一层遍历，s的每个位置是否可分成字典元素
		{
			for (int j = i - 1; j >= 0; j--)
			{
				if (dp[j]) //j之前的元素可以分成字典元素
				{
					if (strset.find(s.substr(j, i - j)) != strset.end())
					{
						dp[i] = true;
						break;
					}
				}
			}
		}

		return dp[s.size()];
	}
};

• bfs
• https://leetcode.com/problems/word-break/discuss/43797/A-solution-using-BFS

bool wordBreak(string s, unordered_set<string> &dict) {
    // BFS
    queue<int> BFS;
    unordered_set<int> visited;
    
    BFS.push(0);
    while(BFS.size() > 0)
    {
        int start = BFS.front();
        BFS.pop();
        if(visited.find(start) == visited.end())
        {
            visited.insert(start);
            for(int j=start; j<s.size(); j++)
            {
                string word(s, start, j-start+1);
                if(dict.find(word) != dict.end())
                {
                    BFS.push(j+1);
                    if(j+1 == s.size())
                        return true;
                }
            }
        }
    }
    
    return false;
}

• dfs
• https://leetcode.com/problems/word-break/discuss/162377/Beats-100-with-C++

class Solution {
public:
    bool dfs(string s, int start, vector<string>& wordDict,vector<bool>& visited){
        if(start == s.size()) return true;
        if(visited[start]) return false;
        visited[start] = true;
        for(string word:wordDict){
            int len = word.size();
            string w = s.substr(start,len);
            int end = start+len;
            if(end>s.size())continue;
            if(w == word&&dfs(s,end,wordDict,visited)){
                return true;
            }
        }
        return false;
    }
    bool wordBreak(string s, vector<string>& wordDict) {
        vector<bool> visited(s.size(),false);
        return dfs(s,0,wordDict,visited);
;    }
};

题目来源

• 139. Word Break