提交了格灵深瞳的简历后，收到需要先进行一个简单的技术测试的通知，临时抱佛脚，先刷刷上面几道题：
题目要求
A zero-indexed array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices, i.e.
A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].
Sum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.
For example, consider the following array A consisting of N = 8 elements:
A[0] = -1
A[1] = 3
A[2] = -4
A[3] = 5
A[4] = 1
A[5] = -6
A[6] = 2
A[7] = 1
P = 1 is an equilibrium index of this array, because:
- A[0] = −1 = A[2] + A[3] + A[4] + A[5] + A[6] + A[7]
P = 3 is an equilibrium index of this array, because:
- A[0] + A[1] + A[2] = −2 = A[4] + A[5] + A[6] + A[7]
P = 7 is also an equilibrium index, because:
- A[0] + A[1] + A[2] + A[3] + A[4] + A[5] + A[6] = 0
and there are no elements with indices greater than 7.
P = 8 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N.
Write a function:
int solution(int A[], int N);
that, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.
For example, given array A shown above, the function may return 1, 3 or 7, as explained above.
Assume that:
- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
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.
代码：
int solution(vector<int> &A){
int sum = 0;
for(int i =0;i<A.size();i++){
sum = sum + A[i];
}
int for_sum = 0;
int result = 0;
for(int i =0;i<A.size();i++){
sum = sum-A[i];
if(sum == for_sum){
result = i;
break;
}
for_sum = for_sum + A[i];
}
return result;
}