TapeEquilibrium
Minimize the value |(A[0] + ... + A[P-1]) - (A[P] + ... + A[N-1])|.
Task description
A non-empty 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:
class Solution { public int solution(int[] A); }
that, given a non-empty 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.
Write an efficient algorithm for the following assumptions:
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].
這題滿容易的，只要先算出總和是多少，再用總和減first part的數字就是second part的數字，然後再進行比大小即可
public static int tapeEquilibrium(int[] a) {
int sum = 0, firstSum = 0, secondSum, anser, temp = Integer.MAX_VALUE;
for (int i : a) {
sum += i;
}
for (int i = 0; i < a.length - 1; i++) {
firstSum += a[i];
secondSum = sum - firstSum;
anser = Math.abs(firstSum - secondSum);
if (temp > anser) {
temp = anser;
}
}
return temp;
}
執行結果：https://app.codility.com/demo/results/trainingD85W9T-KQT/
Cheng® 