TapeEquilibrium

• 初始化 head = A[0] 和 tail = accumulate(A.begin() + 1, A.end(), 0)
• 再用一個迴圈加一個元素、減一個元素，比 minDiff 的大小

Solution

#include <bits/stdc++.h>
using namespace std;

int solution(vector<int> &A)
{
    int n = A.size();
    int head = A[0];
    int tail = accumulate(A.begin() + 1, A.end(), 0);

    int minDiff = abs(head - tail);

    for (int i = 1; i < n - 1; ++i)
    {
        head += A[i];
        tail -= A[i];
        int cur = abs(head - tail);
        minDiff = min(minDiff, cur);
    }
    return minDiff;
}

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(1)$