leetcode -- word break

 一个没有把百酒都尝遍的人，是不会体会到清水之味的~

[问题描述]

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".

 

[解题思路]

DP: ans[i]表示s的下标0~i子串是否可以被dict表示

ans[i] = 存在k满足ans[k]&dict.contains(s.substr(k+1, i-k))==true即可. 

 

bool Solution::wordBreak(std::string s, std::tr1::unordered_set<std::string> &dict)
{
    if (s.length() == 0 || dict.size()==0)
        return false;
    int length = s.length();
    bool ans[length];
    for (int i = 0; i < length; i++)
        ans[i] = false;
    for (int i = 0; i < length; i++){
        if (dict.find(s.substr(0, i+1)) != dict.end())
            ans[i] = true;
        for (int j = 0; j < i; j++)
            if (ans[j]&&dict.find(s.substr(j+1, i-j))!=dict.end()){
                ans[i] = true;
                break;
            }
    }
    return ans[length-1];
}