在一个数组算法中实现K个堆栈 [英] Implementing K stacks in one array algorithm

问题描述

如何以最佳存储使用率在阵列中实现K个堆栈(堆栈应该是动态的)?

How to implement K stacks in an array, with best storage usage (stacks should be dynamic)?

推荐答案

好吧，如果您只担心空间使用情况，而不必担心堆栈操作会占用 O(N)，您可以使用阵列的前几个单元格来管理堆栈:

Well, if you're only worried about space usage, and don't care that stack operations can take O(N), you can use the array's first few cells to manage the stacks:

Array [0] -堆栈0的末尾

Array [1] -堆栈1的末尾

...

Array [K-1] =堆栈K的末尾

Stack n 开始于 Array [n-1] ，结束于 Array [n] (独占-[Array [n-1]，Array [n])).如果Array [n-1] == Array [n] ，则堆栈为空.第一个堆栈从K开始，因此第一个 Array [0] .. Array [K-1] = K

Stack n starts at Array[n-1] and ends at Array[n] (exclusive - [Array[n-1], Array[n]) ). If Array[n-1]==Array[n] the stack is empty. The first stack starts at K, so at first Array[0]..Array[K-1] = K

当推入堆栈时，只需将其下面堆栈中的所有元素移动，并分别调整指针.

When you push into a stack, just move all the elements in the stacks below it, and adjust the pointers respectively.

它将为您提供所需的内存限制.

It'll get you the memory constraint you need.

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