Leetcode--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”.

思路

S能拆成功的话，说明
s.substring(0,i)能拆成功，然后 s.substring(i)是一个在字典中的单词。
后者是一步check: dict.contains(s.substring(i));
前者是需要记录的信息dp[i]表示可拆
然后从头撸一遍就行了

代码

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