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).
原文：http://www.cnblogs.com/cane/p/3973828.html