C++训练3

题目1：

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(int A[], int N);

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).

程序：

#include <iostream>
#include <vector>
using namespace std;
int solution1(vector<int> &A)
{
	int diff=0,min_diff=5000,sum1=0,sum2=0;
	auto v=A;
	auto it=v.begin();
	sum1=*it;
	it++;
	while(it!=v.end()) {sum2+=*it;it++;}
	diff= abs(sum1 - sum2);
	min_diff=diff;
	it=v.begin()+1;//重置
	while (it != (v.end()-1))
	{
		sum1+=*it;
		sum2-=*it;
		diff = abs(sum1 - sum2);
		if(diff<min_diff)
			min_diff=diff;
		it++;
	}
	return min_diff;
}
int solution(vector<int> &A)
{
	if(A.empty())
		return -1; 
	else
		return solution1(A);
}
int main()
{
	vector<int> v={-1000,1000};
	cout<<solution(v)<<endl;
	getchar();
}

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