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LeetCode:Word Break（DP）

题目地址：http://oj.leetcode.com/problems/word-break/

简单的动态规划问题，采用自顶向下的备忘录方法，代码如下：

 1 class Solution {
 2 public:
 3     bool dictContain(unordered_set<string> &dict, string s)
 4     {
 5         unordered_set<string>::iterator ite = dict.find(s);
 6         if(ite != dict.end())
 7             return true;
 8         else return false;
 9     }
10 
11     bool wordBreak(string s, unordered_set<string> &dict)
12     {
13         // Note: The Solution object is instantiated only once and is reused by each test case.
14         if(dict.empty())
15             return false;
16         const int len = s.size();
17         bool canBreak[len]; //canBreak[i] = true 表示s[0~i]是否能break
18         memset(canBreak, 0, sizeof(bool)*len);
19         for(int i = 1; i <= len; i++)
20         {
21             if(canBreak[i-1] == false && dictContain(dict, s.substr(0, i)))
22                 canBreak[i-1] = true;
23 
24             if(canBreak[i-1] == true)
25             {
26                 if(i == len)return true;
27                 for(int j = 1; j <= len - i; j++)
28                 {
29                     if(canBreak[j+i-1] == false && dictContain(dict,s.substr(i, j)))
30                         canBreak[j+i-1] = true;
31 
32                     if(j == len - i && canBreak[j+i-1] == true)return true;
33 
34                 }
35             }
36 
37         }
38 
39         return false;
40     }
41 
42 
43 };

注意：在本机调试时，编译器要开启c++11支持，因为#include<unordered_set>是c++11的标准