*[codility]GenomicRangeQuery
http://codility.com/demo/take-sample-test/genomicrangequery
这题有点意思。
一开始以为是RMQ或者线段树,但这样要O(n*logn)。考虑到只有四种字符,可以用数组记录每个字符i之前出现过几次。
二,查询区间是闭区间,所以要处理off by one的问题。
// you can also use includes, for example: // #include <algorithm> vector<int> solution(string &S, vector<int> &P, vector<int> &Q) { // write your code in C++98 int len = S.length(); vector<int> A(len+1); vector<int> C(len+1); vector<int> G(len+1); vector<int> T(len+1); A[0] = 0; C[0] = 0; G[0] = 0; T[0] = 0; for (int i = 1; i <= len; i++) { A[i] = A[i-1]; C[i] = C[i-1]; G[i] = G[i-1]; T[i] = T[i-1]; if (S[i-1] == 'A') { A[i]++; } else if (S[i-1] == 'C') { C[i]++; } else if (S[i-1] == 'G') { G[i]++; } else if (S[i-1] == 'T') { T[i]++; } } vector<int> ans; for (int i = 0; i < P.size(); i++) { int p = P[i]; int q = Q[i] + 1; if (A[q] - A[p] > 0) ans.push_back(1); else if (C[q] - C[p] > 0) ans.push_back(2); else if (G[q] - G[p] > 0) ans.push_back(3); else if (T[q] - T[p] > 0) ans.push_back(4); } return ans; }