TapeEquilibrium (Codility)

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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;
}

TapeEquilibrium (Codility)

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原文地址：http://www.cnblogs.com/oboorange/p/3963428.html