LeetCode: Word Break I && II

I

title:

https://leetcode.com/problems/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".

思路：一开始就考虑直接使用dfs超时了。。。

class Solution {
public:
    bool wordBreak(string s, unordered_set<string>& wordDict) {
        if (s.size() == )
            return true;
        string tmp;
        for (int i = ; i < s.size(); i++){
            tmp.push_back(s[i]);
            if (wordDict.find(tmp) != wordDict.end()){
                string tmp1(s.begin()+i+,s.end());
                if (wordBreak(tmp1,wordDict))
                    return true;
            }
        }
        return false;
    }
};

其实对于这样的问题，是可以想到是可以使用动态规划的。

一个DP问题。定义possible[i] 为S字符串上[0,i]的子串是否可以被segmented by dictionary.

那么

possible[i] = true      if  S[0,i]在dictionary里面

= true      if   possible[k] == true 并且 S[k+1,j]在dictionary里面， 0<k<i

= false      if    no such k exist.

class Solution {
public:
    bool wordBreak(string s, unordered_set<string>& wordDict) {
        int len = s.size();
        vector<bool> dp(len+,false);
        dp[] = true;
        for (int i = ; i < len+; i++){
            for (int k = ; k < i; k++){
                if (dp[k] && wordDict.find(s.substr(k,i-k)) != wordDict.end()){
                    dp[i] = true;
                    break;//break是因为题目只是需要判断是否存在
                }
            }
        }
        return dp[len];
    }
};

II

https://leetcode.com/problems/word-break-ii/

title:

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”].

思路：使用dp，并且保存中间结果。注意，在I中找到一个解就直接break，这里需要记录下这个信息。另外prev[i][j]为true表示字串[j,i]在字典中

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string>& wordDict) {        int len = s.size();
        vector<bool> dp(len+,false);
        vector<vector<bool> > prev(len+,vector<bool>(len,false));
        dp[] = true;
        for (int i = ; i < len+; i++){
            for (int k = ; k < i; k++){
                if (dp[k] && wordDict.find(s.substr(k,i-k)) != wordDict.end()){
                    dp[i] = true;
                    prev[i][k] = true;
                }
            }
        }
        vector<string> result;
        vector<string> results;
        genPath(len,s,dp,prev,result,results);
        return results;
    }    void genPath(int cur,string s, vector<bool> dp,vector<vector<bool> > prev, vector<string> &result,vector<string> &results){
        if (cur == ){
            string tmp;
            for (int i = result.size()-; i > ; i--){
                tmp += result[i] + " ";
            }
            tmp += result[];
            results.push_back(tmp);
            return ;
        }
        for (int i = ; i  < s.size(); i++){
            if (prev[cur][i]){
                result.push_back(s.substr(i,cur-i));
                genPath(i,s,dp,prev,result,results);
                result.pop_back();
            }
        }
    }
};