【LeetCode】139. Word Break

Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

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

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

 

bpos表示从字符串是否可以从开头cut到当前位置。

bpos[i]==true代表从字符串开头到第i个字符结束，存在合理的cut。

在字符串s前面加上0，从而为动态规划设置初值即bpos[0]==true

 

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