Department of Computer Science and Software Engineering
被引用18|浏览2
摘要
The maximum subarray problem for a one- or two-dimensional array is to find the array portion that maiximizes the sum of array elements in it. The K-maximum subarray problem is to find the K subarrays with largest sums. We improve the time complexity for the one-dimensional case from $O(min\{K+n\log^2 n, n\sqrt{K}\})$ for 0 ≤ K ≤ n(n–1)/2 to O(nlog K + K2) for K ≤ n. The latter is better when $K \le \sqrt n\log n$. If we simply extend this result to the two-dimensional case, we will have the complexity of O(n3log K + K2n2). We improve this complexity to O(n3) for $K \le \sqrt{n}$.