思路：统计每一个字符前面的四个字符的个数(利用前缀和数组)，这样就能在O（1）时间得到某个区间某个字符的个数

开始想到的是O（n^2）的空间，这个思路比较好

 1 // you can use includes, for example: 2 // #include <algorithm> 3  4 // you can write to stdout for debugging purposes, e.g. 5 // cout << "this is a debug message" << endl; 6  7 vector<int> solution(string &S, vector<int> &P, vector<int> &Q) { 8     // write your code in C++11 9     int len = S.length();10     vector<int> res;11     vector<vector<int> > arr(len);12     13     for(int k = 0 ; k < len ; ++k)14     {15         arr[k].resize(4);16     }17     int i,j;18     for(i = 0; i < len; i++)19     {20      char c = S[i];21      if(c == ‘A‘) arr[i][0] = 1;22      if(c == ‘C‘) arr[i][1] = 1;23      if(c == ‘G‘) arr[i][2] = 1;24      if(c == ‘T‘) arr[i][3] = 1;25    }26     for(i = 1 ; i < len ; ++i)27     {28         for(j = 0 ; j < 4 ; ++j)29         {30             arr[i][j] += arr[i-1][j];31         }32     }33     for(i = 0 ; i < P.size() ; ++i)34     {35         int x = P[i];36         int y = Q[i];37         for(j = 0 ; j < 4 ; ++j)38         {39             int sub = 0;40             if(x-1>=0)41             {42                 sub = arr[x-1][j];43             }44             if(arr[y][j] - sub > 0)45             {46                 res.push_back(j+1);47                 break;48             }49         }50     }51     return res;52 }

A DNA sequence can be represented as a string consisting of the letters A, C, G and T, which correspond to the types of successive nucleotides in the sequence. Each nucleotide has an impact factor, which is an integer. Nucleotides of types A, C, G and T have impact factors of 1, 2, 3 and 4, respectively. You are going to answer several queries of the form: What is the minimal impact factor of nucleotides contained in a particular part of the given DNA sequence?

The DNA sequence is given as a non-empty string S =S[0]S[1]...S[N-1] consisting of N characters. There are M queries, which are given in non-empty arrays P and Q, each consisting of M integers. The K-th query (0 ≤ K < M) requires you to find the minimal impact factor of nucleotides contained in the DNA sequence between positions P[K] and Q[K] (inclusive).

For example, consider string S = CAGCCTA and arrays P, Q such that:

    P[0] = 2    Q[0] = 4    P[1] = 5    Q[1] = 5    P[2] = 0    Q[2] = 6

The answers to these M = 3 queries are as follows:

• The part of the DNA between positions 2 and 4 contains nucleotides G and C (twice), whose impact factors are 3 and 2 respectively, so the answer is 2.
• The part between positions 5 and 5 contains a single nucleotide T, whose impact factor is 4, so the answer is 4.
• The part between positions 0 and 6 (the whole string) contains all nucleotides, in particular nucleotide Awhose impact factor is 1, so the answer is 1.

Write a function:

vector<int> solution(string &S, vector<int> &P, vector<int> &Q);

that, given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.

The sequence should be returned as:

• a Results structure (in C), or
• a vector of integers (in C++), or
• a Results record (in Pascal), or
• an array of integers (in any other programming language).

For example, given the string S = CAGCCTA and arrays P, Q such that:

    P[0] = 2    Q[0] = 4    P[1] = 5    Q[1] = 5    P[2] = 0    Q[2] = 6

the function should return the values [2, 4, 1], as explained above.

Assume that:

• N is an integer within the range [1..100,000];
• M is an integer within the range [1..50,000];
• each element of arrays P, Q is an integer within the range [0..N − 1];
• P[K] ≤ Q[K], where 0 ≤ K < M;
• string S consists only of upper-case English letters A, C, G, T.

Complexity:

• expected worst-case time complexity is O(N+M);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Colidity--GenomicRangeQuery