LeetCode ||& Word Break && Word Break II（转）——动态规划

一、

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".

 1 class Solution {
 2 public:
 3     bool wordBreak(string s, unordered_set<string> &dict) {
 4         int len=s.length();
 5         vector<bool> v(len+1,false);
 6         v[0]=true;
 7         for(int pos=0;pos<len;pos++){
 8             for(int i=pos;v[pos]&&i<len;i++){
 9                 if(dict.find(s.substr(pos,i-pos+1))!=dict.end())
10                     v[i+1]=true;
11             }
12         }
13         return v[len];
14     }
15 };

二、

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"].

即不仅要确定字符串是否能被字典分割，还要找出所有可能的组合。参考word break那题的DP思路，首先，从尾部开始逆向看字符串 s ，循环截取一个存在的词（milestone 1），然后在截取的位置递归，继续向前看，继续截取。。。知道到达头部，此时组合出一种答案；然后进入milestone 1 处的下一次循环，如下图的milestone 1‘，截取另外一个词，找另外一个答案。。。

代码如下：

 1 class Solution {  
 2     vector<string> midres;  
 3     vector<string> res;  
 4     vector<bool> *dp;  
 5 public:  
 6     vector<string> wordBreak(string s, unordered_set<string> &dict) {  
 7         int len = s.length();  
 8           
 9         dp = new vector<bool>[len];  
10         for(int i=0; i<len; ++i){  
11             for(int j=i; j<len; ++j){  
12                 if(dict.find(s.substr(i, j-i+1))!=dict.end()){  
13                     dp[i].push_back(true);      //第二维的下标实际是:单词长度-1  
14                 }else{  
15                     dp[i].push_back(false);     //数组第二维用vector,size不一定是n,这样比n*n节省空间  
16                 }  
17             }  
18         }  
19         func(s, len-1);  
20         return res;  
21     }  
22       
23     void func(const string &s, int i){  
24         if(i>=0){  
25             for(int j=0; j<=i; ++j){  
26                   
27                 if(dp[j][i-j]){ //注意此处的第二个下标是 i-j，不是i,因为数组的第二维长度是不固定的,第二维的下标实际是单词长度-1  
28                   
29                     midres.push_back(s.substr(j, i-j+1));  
30                     func(s, j-1);  
31                     midres.pop_back();  //继续考虑for循环的下一个分段处  
32                 }  
33             }  
34             return;  
35         }  
36         else{  
37             string str;  
38             for(int k=midres.size()-1; k>=0; --k){  //注意遍历的顺序是倒序的  
39                 str += midres[k];   //注意此处是k,不是i  
40                 if(k>0)  
41                     str += " ";  
42             }  
43             res.push_back(str);  
44             return;  
45         }  
46     }  
47 };