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Word Break

时间:2015-06-21 18:23:28      阅读:116      评论:0      收藏:0      [点我收藏+]

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Description:

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".

Code:

 1   bool wordBreak(string s, unordered_set<string>& wordDict) {
 2         //flag[i][j]表示s[i]...s[j]是否为一个word
 3       int length = s.size();  
 4      bool flag[MAX][MAX]={false};
 5       for (int i = 0; i < length; ++i)
 6       {
 7         string str(s,i,1);
 8         if (wordDict.find(str)!=wordDict.end())
 9             flag[i][i] = true;
10       }
11      
12      for (int r=2; r<=length; ++r)
13      {//计算长度为2,3,...n的字符串
14          for (int i=0; i <= length-r; ++i)
15          {
16              int j = i+r-1;//j<=length-1
17             //计算flag[i][j];
18             string str(s,i,r);
19             if (wordDict.find(str)!=wordDict.end())
20             {
21                 flag[i][j] = true;
22                 continue;
23             }
24              for (int k = i; k < j; ++k)
25              {
26                  if (flag[i][k]&&flag[k+1][j])
27                  {
28                     flag[i][j] = true;
29                     break;
30                  }
31              }
32          }
33      }
34       return flag[0][length-1];
35     }

 

Word Break

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原文地址:http://www.cnblogs.com/happygirl-zjj/p/4592163.html

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