Leetcode:Word Break 字符串分解为单词

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".

解题分析：注意词典中的每个单词可以多用，比如 s = "bb"  dict = ["a", "b"]需要返回true

class Solution {

public:

    bool wordBreak(string s, unordered_set<string> &dict) {

        std::vector<bool> dp(s.size() + 1, false);  

        dp[0] = true;

        for (int i = 1; i <= s.size(); ++i) {

            for (int j = i - 1; j >= 0; --j) {

                std::string str = s.substr(j, i - j);

                if (dp.at(j) == true && dict.find(str) != dict.end()) {

                    dp.at(i) = true;

                }

            }

        }

        return dp.at(s.size());

    }

};

 

Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.



Return all such possible sentences.



For example, given

s = "catsanddog",

dict = ["cat", "cats", "and", "sand", "dog"].



A solution is ["cats and dog", "cat sand dog"].