139. Word Break

问题描述

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.

题目链接：

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思路分析

给一个string和一组word，判断string是否是由这些word组成的。

动态规划的思想，判断i之前的j个字符组成的word是否在字典中出现，看最后一个字符位置是否是true即可。

代码

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        if(wordDict.size()==0) return false;

        vector<bool> dp(s.size()+1,false);
        dp[0]=true;
        int prev = 0;
        for(int i=1;i<=s.size();i++)
        {
            for(int j=i-1;j>=0;j--)
            {
                if(dp[j])
                {
                    string word = s.substr(j,i-j);
                    if (std::find(wordDict.begin(), wordDict.end(), word) != wordDict.end())
                    {
                        dp[i]=true;
                        break; //next i
                    }
                }
            }
        }

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

时间复杂度：$O(n^2)$
空间复杂度：$O(n)$

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反思

一开始好无头绪，后来又了一丢丢思路。