Substring with Concatenation of All Words

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

思路：

设L中每个字符长度为l，index[i]用来存S中以第i个字符开头且长度为l的字符串是否包含在L中，如果在则为L的值，否则是-1。这里需要注意的是L中的字符串可能会有重复的，所以在遇到index[i]不为-1的情况就不继续往后算。这一步的复杂度是O(mnl)(m为S的长度，n为L中字符串的数量，l为L中每个字符串的长度)。

needNum记录L中每个字符串需要的次数，hasNum记录当前找到的L中字符串的次数。

对S进行一次遍历，如果index[i]等于-1继续往后走，否则的话计算从i开始数n个字符后中hasNum的值。最后比较hasNum和needNum，如果全部相等，则把该结果存入最终vector。

代码：

 1     vector<int> findSubstring(string S, vector<string> &L) {
 2         int ls = S.length(), n = L.size();
 3         vector<int> result;
 4         if(n == 0)
 5             return result;
 6         int ll = L[0].length();
 7         int index[ls];
 8         memset(index, -1, sizeof(int)*ls);
 9         int needNum[n], hasNum[n];
10         for(int i = 0; i < n; i++)
11             needNum[i] = 1;
12         int i, j, k;
13         const char *s = S.c_str();
14         for(i = 0; i < n; i++){
15             const char *l = L[i].c_str();
16             char *tmp = strstr(s, l);
17             if(tmp != NULL){
18                 if(index[tmp-s] != -1){
19                     needNum[index[tmp-s]]++;
20                     needNum[i] = 0;
21                     continue;
22                 }
23             }
24             while(tmp){
25                 index[tmp-s] = i;
26                 tmp = strstr(tmp+1, l);
27             }
28         }
29         for(i = 0; i < ls; i++){
30             if(index[i] == -1)
31                 continue;
32             memset(hasNum, 0, sizeof(int)*n);
33             for(j = i; j < (ls<(i+n*ll)?ls:(i+n*ll)); j+=ll){
34                 if(needNum[index[j]] == hasNum[index[j]]){
35                     break;
36                 }
37                 hasNum[index[j]]++;
38             }
39             for(k = 0; k < n; k++){
40                 if(needNum[k] != -1 && hasNum[k] != needNum[k])
41                     break;
42             }
43             if(k == n)
44                 result.push_back(i);
45         }
46         return result;
47     }

 第二遍：

 1         vector<int> findSubstring(string S, vector<string> &L) {  
 2             // Start typing your C/C++ solution below  
 3             // DO NOT write int main() function  
 4             map<string, int> words;  
 5             map<string, int> count;  
 6             vector<int> result;  
 7             int wordNum = L.size();  
 8             if (wordNum == 0) return result;  
 9             for (int i = 0; i < wordNum; ++i)  
10                 ++words[L[i]];  
11               
12             int wordSize = L[0].size();  
13             int slength = S.size();  
14             for (int i = 0; i <= slength - wordSize*wordNum; ++i)  
15             {  
16                 count.clear();  
17                 int j = 0;  
18                 for (; j < wordNum; ++j)  
19                 {  
20                     string w = S.substr(i+j*wordSize, wordSize);   
21                     if(words.find(w) == words.end())    
22                         break;    
23                     ++count[w];    
24                     if(count[w] > words[w])    
25                         break;    
26                 }  
27                 if (j == wordNum) result.push_back(i);  
28             }  
29             return result;  
30         }

 

posted on 2013-12-03 21:36  waruzhi  阅读(231)  评论(0)  编辑  收藏  举报

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