1.题目（HARD）
A non-empty zero-indexed array A of N integers is given. A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. Thesum of a slice (P, Q) is the total of A[P] + A[P+1] + ... + A[Q].
A min abs slice is a slice whose absolute sum is minimal.
For example, array A such that:
A[0] = 2 A[1] = -4 A[2] = 6 A[3] = -3 A[4] = 9
contains the following slices, among others:
- (0, 1), whose absolute sum = |2 + (−4)| = 2
- (0, 2), whose absolute sum = |2 + (−4) + 6| = 4
- (0, 3), whose absolute sum = |2 + (−4) + 6 + (−3)| = 1
- (1, 3), whose absolute sum = |(−4) + 6 + (−3)| = 1
- (1, 4), whose absolute sum = |(−4) + 6 + (−3) + 9| = 8
- (4, 4), whose absolute sum = |9| = 9
Both slices (0, 3) and (1, 3) are min abs slices and their absolute sum equals 1.
Write a function:
int solution(int A[], int N);
that, given a non-empty zero-indexed array A consisting of N integers, returns the absolute sum of min abs slice.
For example, given:
A[0] = 2 A[1] = -4 A[2] = 6 A[3] = -3 A[4] = 9
the function should return 1, as explained above.
Assume that:
- N is an integer within the range [1..100,000];
- each element of array A is an integer within the range [−10,000..10,000].
Complexity:
- expected worst-case time complexity is O(N*log(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.
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2.题目分析
这个题目看上去很简单。
slice求和，prefix的经典解法。不过，这个要求的是和的绝对值。这就有一点生硬了。贪心算法或者蠕虫都不太实用。
有点着急呢。。眼看答案就在眼前，可是就是想不出。过了20分钟，我有点开窍了。
首先是看到了时间复杂度为O（N*logN）.这感觉是要排序啊，不过对元数据排序貌似没有用啊。突然意识到，可以对prefix排序。
瞬间感觉自己很牛逼。
你想，prefix任何两者的差，即是这一段slice的和，那么我们现在其实找的就是这段和的最小绝对值。我们将prefix排序以后，那么就是大小最相近的两个prefix的值会成为相邻。对其做差，就是找到一段最小值。我们顺序遍历这个prefix数组，找到所有的相邻项相差的最小值，不就是我们要找的那个绝对值吗？(因为已经从大到小排序了)。
当然，还是要做一些其他的处理，比如找到prefix的绝对值最小值，和数组的绝对值的最小值，同得出的结论进行比较。
就必然得出了最后的答案。