vector<int> solution(int N, vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
int max = 0;
int lastmax = 0;
vector<int>res(N);
for(int i : A)
{
if(i>=1&& i<=N)
{
if( res[i-1] < lastmax )
{
res[i-1] = lastmax + 1;
}
else
{
res[i-1] ++;
}
if(res[i-1] > max)
{
max = res[i-1];
}
}
else if(i == N+1)
{
lastmax = max;
}
}
for(int i = 0; i < N; i++)
{
if(res[i] < lastmax )
{
res[i] = lastmax;
}
}
return res;
}
问题描述
Task description
You are given N counters, initially set to 0, and you have two possible operations on them:
- increase(X) − counter X is increased by 1,
- max counter − all counters are set to the maximum value of any counter.
A non-empty array A of M integers is given. This array represents consecutive operations:
- if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
- if A[K] = N + 1 then operation K is max counter.
For example, given integer N = 5 and array A such that:
A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4
the values of the counters after each consecutive operation will be:
(0, 0, 1, 0, 0) (0, 0, 1, 1, 0) (0, 0, 1, 2, 0) (2, 2, 2, 2, 2) (3, 2, 2, 2, 2) (3, 2, 2, 3, 2) (3, 2, 2, 4, 2)
The goal is to calculate the value of every counter after all operations.
Write a function:
vector<int> solution(int N, vector<int> &A);
that, given an integer N and a non-empty array A consisting of M integers, returns a sequence of integers representing the values of the counters.
Result array should be returned as a vector of integers.
For example, given:
A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4
the function should return [3, 2, 2, 4, 2], as explained above.
Write an efficient algorithm for the following assumptions:
- N and M are integers within the range [1..100,000];
- each element of array A is an integer within the range [1..N + 1].