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).
这题还是挺复杂的，最主要设计到两大块代码，一个是hash字符串，还有一个是找最小窗口。可以做到O(n)吧，n = s.length()
1 class Solution { 2 private: 3 int count[1000]; 4 int countSize; 5 map<string, int> index; 6 vector<int> ret; 7 public: 8 vector<int> findSubstring(string S, vector<string> &L) { 9 // Start typing your C/C++ solution below 10 // DO NOT write int main() function 11 ret.clear(); 12 if (L.size() == 0) 13 return ret; 14 15 index.clear(); 16 countSize = 0; 17 for(int i = 0; i < L.size(); i++) 18 if (index.count(L[i]) > 0) 19 count[index[L[i]]]++; 20 else 21 { 22 index[L[i]] = countSize; 23 count[countSize++] = 1; 24 } 25 26 int step = L[0].size(); 27 28 vector<int> a; 29 30 for(int i = 0; i < step; i++) 31 { 32 a.clear(); 33 for(int j = i; j < S.size(); j += step) 34 { 35 if (j + step <= S.size()) 36 { 37 string s(S, j, step); 38 if (index.count(s) > 0) 39 a.push_back(index[s]); 40 else 41 a.push_back(-1); 42 } 43 } 44 45 int beg = -1; 46 int end = 0; 47 int size = L.size(); 48 while(end < a.size()) 49 { 50 if (a[end] != -1) 51 { 52 if (count[a[end]] > 0) 53 { 54 if (beg == -1) 55 beg = end; 56 size--; 57 count[a[end]]--; 58 } 59 else 60 { 61 while(beg < end) 62 { 63 count[a[beg]]++; 64 size++; 65 if (a[beg++] == a[end]) 66 break; 67 } 68 count[a[end]]--; 69 size--; 70 } 71 } 72 else 73 { 74 size = L.size(); 75 if (beg != -1) 76 { 77 for(int i = beg; i < end; i++) 78 count[a[i]]++; 79 } 80 beg = -1; 81 } 82 83 end++; 84 85 if (size == 0) 86 { 87 ret.push_back(beg * step + i); 88 size++; 89 count[a[beg]]++; 90 beg++; 91 } 92 } 93 94 if (beg != -1) 95 { 96 for(int i = beg; i < end; i++) 97 count[a[i]]++; 98 } 99 } 100 101 return ret; 102 } 103 };