Codility-task 1-Tape Equilibrium

Codility是个great的OJ，用起来非常爽，所以我要来CSDN上安利一下！


网址是 https://codility.com/programmers/challenges/

Task Description

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.

题目的等级是easy，无需多言，下边是我的解答。

//Codility 
//Zhang Wenjian 07-06-2015
int solution(vector<int> &A) {
    // write your code in C++11
    long long sum;
    long long th;
    
    for(vector<int>::size_type it=0;it<A.size();it++)
    sum+=A[it];
    th=abs(sum-2*A[0]);
    if(A.size()==2)
    return abs(A[0]-A[1]);
    for(vector<int>::size_type it=0;it<A.size()-1;it++)
    {
        sum-=2*A[it];
        if(abs(sum)<th)
        th=abs(sum);
    }
    
    return th;
}