[LeetCode] 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".

 

方法一：

DFS，

start已知，当start超过str长度时，说明全部字符串都能找到了。。

小数据可过，大数据时超时

Submission Result: Time Limit Exceeded

Last executed input:

"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]

class Solution {
    public:
        bool wordBreak(string s, unordered_set<string> &dict)
        {
            return wordBreak(s,0, dict);
        }   

        bool wordBreak(string s, int start, unordered_set<string> & dict)
        {   
            size_t size =  s.size();
            if(size == 0) return false;
            if(start >= size)
                return true;

            for( int i = start; i <size; i++)
            {   
                string str = s.substr(start, i-start+1);
                if(dict.find(str) != dict.end())
                {   
                    if(wordBreak(s,i+1, dict))
                       return true;
                }   

            }
            return false;
        }
};

 

方法二：

DP

设状态为 f(i)，表示 s[0,i] 是否可以分词，则状态转移方程为
f(i) = any_of(f(j)&&s[j + 1, i] 2 dict), 0  j < i

 

bool wordBreak2(string s, set<string> &dict) {
    vector<bool> f(s.size() + 1, false);
    f[0] = true; 
    for (int i = 1; i <= s.size(); ++i) {
        for (int j = i - 1; j >= 0; --j) {
            if (f[j] && dict.find(s.substr(j, i - j)) != dict.end()) {
                f[i] = true;
                break;
            }   
        }   
    }   
    return f[s.size()];
}