[LeetCode] 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".

  1 #include <iostream>
  2 #include <string>
  3 #include <set>
  4 #include <cstdio>
  5 
  6 using namespace std;
  7 
  8 set<string> myset;
  9 bool ** pmatrix = NULL;
 10 
 11 template<typename T>
 12 void BuildMatrix(T *** pmaze,unsigned row_num,unsigned column_num)
 13 {
 14     *pmaze = new T*[row_num];
 15     for(unsigned i=0;i<row_num;++i){
 16         (*pmaze)[i] = new T[column_num];
 17     }
 18 }
 19 
 20 template<typename T>
 21 void ReleaseMatrix(T ***pmaze,unsigned row_num)
 22 {
 23     if(!pmaze) return;
 24 
 25     for(unsigned i=0;i<row_num;++i){
 26         delete [](*pmaze)[i];
 27     }
 28 
 29     delete [](*pmaze);
 30 }
 31 
 32 bool FindWord(int length){
 33     int i=0,j=0;
 34     for(j=i;j<length;++j){
 35         if(pmatrix[i][j]){
 36             if(j==length-1) return true;
 37             else i = j+1;
 38         }
 39     }
 40 
 41     return false;
 42 }
 43 
 44 void Solve(){
 45     myset.clear();
 46 
 47     string strtomatch;
 48     int dicwordnum = 0;
 49     cin>>dicwordnum;
 50 
 51     for(int i=0;i<dicwordnum;++i){
 52         cin>>strtomatch;
 53         myset.insert(strtomatch);
 54     }
 55 
 56     cin>>strtomatch;
 57     unsigned strlength = strtomatch.length();
 58 
 59     BuildMatrix(&pmatrix,strlength,strlength);
 60 
 61     for(int i=0;i<strlength;++i){
 62         for(int j=i;j<strlength;++j){
 63             pmatrix[i][j] = false;
 64         }
 65     }
 66 
 67     for(unsigned i=0;i<strlength;++i){
 68         for(unsigned j=i;j<strlength;++j){
 69             string strdata;
 70             strdata.assign(strtomatch,i,j-i+1);
 71             if(myset.find(strdata)!=myset.end()){
 72                 pmatrix[i][j] = true;
 73             }
 74         }
 75     }
 76 
 77     for(int i=strlength-1;i>=0;--i){
 78         bool isallfalse = true;
 79         for(int j = i;j<strlength;++j){
 80             if(pmatrix[i][j]){
 81                 isallfalse = false;
 82                 break;
 83             }
 84         }
 85 
 86         if(isallfalse&&(i-1>=0)){
 87             for(int j = 0;j<strlength;++j){
 88                 pmatrix[j][i-1] = false;
 89             }
 90         }
 91     }
 92 
 93     if(FindWord(strlength)) cout<<"true"<<endl;
 94     else cout<<"false"<<endl;
 95 
 96     ReleaseMatrix(&pmatrix,strtomatch.length());
 97 }
 98 
 99 int main()
100 {
101     freopen("txt.in","r",stdin);
102     freopen("txt.out","w",stdout);
103 
104     int case_num = 0;
105     cin>>case_num;
106 
107     for(int i=0;i<case_num;++i){
108         cout<<"Case "<<i+1<<": ";
109         Solve();
110     }
111 
112     return 0;
113 }

txt.in

7
1
a a
2
what is whats
3
hello nice to helloto
5
Kobe Carter LeBron Jordan Iverson Yao
4
ab c abcd e abcde
3
car ca rs cars
2
aaaa aa aaaaaaa

txt.out

Case 1: true
Case 2: false
Case 3: true
Case 4: false
Case 5: true
Case 6: true
Case 7: false
posted @ 2014-10-04 23:10  zhouyoulie  阅读(135)  评论(0)  编辑  收藏  举报