1. MaxCounters 计数器 Calculate the values of counters after applying all alternating operations: increase counter by 1; set value of all counters to current maximum.

package com.code;

import java.util.Arrays;

public class Test04_4 {
    public static int[] solution(int N, int[] A) {
        // write your code in Java SE 8
        int size = A.length;
        int [] res = new int[N];
        int max = 0;
        for(int i=0;i<size;i++){
            if(A[i]==N+1){
                if(i>1 && A[i]==A[i-1]){ // handle {max,max,max,max} array
                    continue;
                }
                for(int j=0,sizeJ=res.length;j<sizeJ;j++){
                    res[j] = max;
                }
            }else{
                res[A[i]-1]=res[A[i]-1]+1;
                max = Math.max(max, res[A[i]-1]);
            }
        }
        return res;
    }
     
    public static void main(String[] args) {
        int [] a = {3,4,4,6,1,4,4};
        System.out.println(Arrays.toString(solution(5, a)));
        int[] b = {6,6,6,6};
        System.out.println(Arrays.toString(solution(5, b)));
        int[] c = {1};
        System.out.println(Arrays.toString(solution(1, c)));
    }
}
/**
1. MaxCounters 计数器
Calculate the values of counters after applying all alternating operations: increase counter by 1; 
set value of all counters to current maximum.

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 zero-indexed 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:

class Solution { public int[] solution(int N, int[] A); }

that, given an integer N and a non-empty zero-indexed array A consisting of M integers, 
returns a sequence of integers representing the values of the counters.

The sequence should be returned as:

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

Assume that:

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].
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).
Elements of input arrays can be modified.

*/