Question

我有两种类型的阵列:x[]和y[ ]。阵列尺寸分别为<代码>n和m。

我们需要找到所订购的<代码>k-th,k <<min(m, n), 即:s ={x[0] + y[0], x[0] + y[0], x[0] + y[0], x x × x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x , x x x[m-1] + [0] y [0], xcode, x x x x[n-1] + [0] +[n-1] + [0] y [0], x x x x x x x x x x x x x x x x x x x-1] +[n-1] + [0] +-1, x x x x x x x x x x x x-1] + [0] +[n] +-1, x x x x x x x x x x x x x x x x x x x x x x x x x x


我目前的方法是多少调和一系列阵列。
{x[0] + y[0], x[0] + y ***, ........................, x[0] + y[m-1]}<>。

http://www.ohchr.org。
............
{x[k-1] + y[0], x[k-1] + y[1], ............, x[k-1] + y[m-1]},
{y[0] + x[0], y[0] + x[1], ............, y[0] + x[n-1]},
{y[1] + x[0], y[1] + x[1], ............, y[1] + x[n-1]},
............
{y[k-1] + x[0], y[k-1] + x[1], ............, y[k-1] + x[n-1]},
and then select the k-th element.
我的猜测是,这个问题应该在与聚合有关的算法中具有一定的意义。 实现这一目标的最快途径是什么?

Answer 1

此处建议采用“@Jetunsaure”算法。计算最后结果所需的最低增加额:

def kth_smallest_sum(x, y, k):
    indexes = [0] * k
    row = 0
    for _ in range(k):
        msum = None
        for r in range(row+2):      # this is safe as k < min(len(x), len(y))
            c = indexes[r]
            tsum = x[r] + y[c]
            if msum is None or tsum < msum:
                minr, msum = r, tsum
        ksum = msum
        indexes[minr] += 1
        row = max(row, minr)
    return ksum

样本使用:

kth_smallest_sum([i * 7 for i in range(10)], [i * 5 for i in range(10)], 8)

产出:

友情链接