139. Word Break

Posted 2020-10-18 小河沟大河沟

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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()];
    }
};

题目来源

• 139. Word Break