A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
For example, consider array A such that:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3
We can split this tape in four places:
- P = 1, difference = |3 − 10| = 7
- P = 2, difference = |4 − 9| = 5
- P = 3, difference = |6 − 7| = 1
- P = 4, difference = |10 − 3| = 7
Write a function:
int solution(vector<int> &A);
that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.
For example, given:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3
the function should return 1, as explained above.
Assume that:
- N is an integer within the range [2..100,000];
- each element of array A is an integer within the range [−1,000..1,000].
Complexity:
- expected worst-case time complexity is O(N);
- 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.
Solution：
two elements和small elements还是wrong answer
int solution(vector<int> &A) {
// write your code in C++11
int min;
int left=0;
int right=0;
int temp=0;
for(int a=0;a<A.size();a++){
right+=A[a];
}
min=abs(right-A[0]-A[0]);
for(int a=0;a<A.size();a++) {
left+=A[a];
right-=A[a];
temp=abs(right-left);
if(temp<min){
min=temp;
}
}
return min;
}