一开始用Integer 怎么都超时，发现用int 才能过。
Equilibrium index of a sequence is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes. For example, in a sequence A:
A[0]=-7 A[1]=1 A[2]=5 A[3]=2 A[4]=-4 A[5]=3 A[6]=0
3 is an equilibrium index, because:
A[0]+A[1]+A[2]=A[4]+A[5]+A[6]
6 is also an equilibrium index, because:
A[0]+A[1]+A[2]+A[3]+A[4]+A[5]=0
(sum of zero elements is zero) 7 is not an equilibrium index, because it is not a valid index of sequence A.
If you still have doubts, this is a precise definition: the integer k is an equilibrium index of a sequence
![$A[0],A[1],/ldots,A[n-1]$](http://g.codility.com/static/task_img/equi/task.en.img1.png){kind=small}
if and only if 
and
![$/sum_{i=0}^{k-1} A[i] = /sum_{i=k+1}^{n-1} A[i]$](http://g.codility.com/static/task_img/equi/task.en.img3.png){kind=small}
.
Assume the sum of zero elements is equal zero. Write a function
int equi(int[] A);
that given a sequence, returns its equilibrium index (any) or -1 if no equilibrium indexes exist. Assume that the sequence may be very long.
Copyright 2009-2010 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
01.
public
boolean
overflag=
false
;
02.
public
int
sum(
int
a,
int
b)
03.
{
04.
int
ret=a+b;
05.
if
(((a>
0
)&&(b>
0
))&&(ret<
0
)) overflag =
true
;
06.
if
(((a<
0
)&&(b<
0
))&&(ret>
0
)) overflag =
true
;
07.
return
ret;
08.
}
09.
public
int
equi (
int
[] A ) {
10.
if
(A.length==
0
)
return
-
1
;
11.
if
(A.length==
1
)
return
0
;
12.
if
(A.length==
2
)
return
-
1
;
13.
int
j=
0
;
14.
int
right=
0
;
15.
int
left=
0
;
16.
int
i=
0
;
17.
for
(i=A.length-
1
;i>
0
;i--)
18.
{
19.
right=sum(right,A[i]);
20.
if
(overflag)
break
;
21.
}
22.
23.
if
((right==
0
)&&(i==
0
))
return
0
;
24.
25.
for
(j=
0
;j<i;j++)
26.
{
27.
left=sum(left,A[j]);
28.
if
(overflag)
return
-
1
;
29.
}
30.
if
(left==right)
return
0
;
31.
j=
0
;
32.
while
(j<A.length-
1
)
33.
{
34.
if
(left==right)
return
j;
35.
left=sum(left,A[j]);
36.
if
(overflag)
return
-
1
;
37.
right=sum(right,-A[j+
1
]);
38.
if
(overflag)
return
-
1
;
39.
j++;
40.
}
41.
42.
if
(left==right)
return
j;
43.
return
-
1
;
44.
45.
}
O(n)
| test | time | result |
|---|---|---|
|
example
Test from the task description |
0.524 s. | OK |
|
extreme_empty
Empty array |
0.784 s. | OK |
| extreme_first | 0.572 s. | OK |
|
extreme_large_numbers
Sequence with extremly large numbers testing arithmetic overflow. |
0.540 s. | OK |
| extreme_last | 0.588 s. | OK |
| extreme_single_zero | 0.504 s. | OK |
|
extreme_sum_0
sequence with sum=0 |
0.684 s. | OK |
| simple | 0.516 s. | OK |
| single_non_zero | 0.756 s. | OK |
|
combinations_of_two
multiple runs, all combinations of {-1,0,1}^2 |
0.516 s. |
WRONG ANSWER
got -1, but equilibrium point exists, for example on position 0 |
|
combinations_of_three
multiple runs, all combinations of {-1,0,1}^3 |
1.112 s. | OK |
| small_pyramid | 1.020 s. | OK |
| large_long_sequence_of_ones | 0.552 s. | OK |
| large_long_sequence_of_minus_ones | 0.536 s. | OK |
| medium_pyramid | 0.808 s. | OK |
|
large_pyramid
Large performance test, O(n^2) solutions should fail. |
0.716 s. | OK |
