An algorithm question-symbol balancing

A sequence composed of N integers A1, A2, and. The symbol of element a [I] in the sequence is defined:
When a [I]> 0, SGN (A [I]) = 1; When a [I] = 0, SGN (A [I]) = 0; when a [I] <0, SGN (A [I]) =-1;
The symbol balancing problem requires the length of the longest symbol balancing segment of the given sequence L, that is:
L = max {J-I + 1}, where 1 = <I <= j <= n, and it meets SGN (A [I]) + SGN (A [I + 1]) +... + SGN (A [J]) = 0 indicates the maximum value of all I in the expression J.
For example, when n = 10, the corresponding sequence is: 1, 1,-1,-2, 0, 1,-1,-1, L = 9.
Algorithm Design:
Given a sequence composed of N integers A1, A2, and an, design an O (n) time algorithm to calculate its longest symbol.
The length of the balancing segment.
Data input:
Input data is provided by the file input.txt. The first row of the file is a positive integer N, and the second row is an integer sequence A1, A2,
Result output:
Output the length of the calculated longest sign balancing section to output.txt.
Input File example output file example
Input.txt output.txt
10 9
1 1-1-2 0 1 3-1 2-1

 

# Include <stdio. h>
# Include <assert. h>

# Define sign (x) = 0? 0: (x)> 0? 1:-1 ))

Int calc (INT data [], int CNT)
{
If (CNT <= 0)
Return 0;

Int * value = new int [CNT + 1];
Int * First = new int [CNT + 1];
Int * Last = new int [CNT + 1];

For (INT I = 0; I <CNT + 1; I ++)
First [I] =-1;

/*
Initialize value, then value [I] represents
Sum of the first ith elements 'sign value.
*/
Value [0] = 0;
For (INT I = 1; I <CNT + 1; I ++)
Value [I] = value [I-1] + sign (data [I-1]);

/*
For each element V in value [0... CNT], first [v]
Denotes the index of V's first appearance, and
Last [v] denotes the last appearance.
*/
For (INT I = 0; I <CNT + 1; I ++)
{
Int v = value [I];
If (first [v] =-1)
First [v] = last [v] = I;
Else
Last [v] = I;
}

Int max = 0;
For (INT I = 0; I <CNT + 1; I ++)
If (first [I]! =-1 & max <last [I]-first [I])
Max = last [I]-first [I];

Delete value;
Delete first;
Delete last;

Return Max;
}

Int main ()
{
Int data1 [] = {1, 1,-1,-2, 0, 1, 3,-1, 2,-1 };
Int data2 [] = {1 };
Int data3 [] = {0 };

Int max1 = calc (data1, sizeof (data1)/sizeof (data1 [0]);
Int max2 = calc (data2, sizeof (data2)/sizeof (data2 [0]);
Int max3 = calc (data3, sizeof (data3)/sizeof (data3 [0]);

Assert (max1 = 9 );
Assert (max2 = 0 );
Assert (max3 = 1 );

Return 0;
}