一道笔试题 体现出的严谨和算法

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



if and only if


and



.

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.

}

nalysis

Detected time complexity:

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