Question
Implement N > 0 stacks using a single array to store all stack data (you may use auxiliary arrays in your stack object, but all of the objects in all of the stacks must be in the same array). No stack should be full unless the entire array is full.
eg.
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N = 3; capacity = 10; Stacks stacks = new Stacks(N, capacity); stacks.put(0, 10); stacks.put(2, 11); stacks.pop(0) = 10; stacks.pop(2) = 11; |
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Solution
How was that problem? You can check out the solution in the video below.
Here is the source code for the solution shown in the video:
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public class Stacks { int[] topOfStack; int[] stackData; int[] nextIndex; int nextAvailable = 0; public Stacks(int numStacks, int capacity) { topOfStack = new int[numStacks]; for (int i = 0; i < topOfStack.length; i++) { topOfStack[i] = -1; } stackData = new int[capacity]; nextIndex = new int[capacity]; for (int i = 0; i < nextIndex.length - 1; i++) { nextIndex[i] = i+1; } nextIndex[nextIndex.length - 1] = -1; } public void push(int stack, int value) { if (stack < 0 || stack >= topOfStack.length) { throw new IndexOutOfBoundsException(); } if (nextAvailable < 0) return; int currentIndex = nextAvailable; nextAvailable = nextIndex[currentIndex]; stackData[currentIndex] = value; nextIndex[currentIndex] = topOfStack[stack]; topOfStack[stack] = currentIndex; } public int pop(int stack) { if (stack < 0 || stack >= topOfStack.length || topOfStack[stack] < 0) { throw new IndexOutOfBoundsException(); } int currentIndex = topOfStack[stack]; int value = stackData[currentIndex]; topOfStack[stack] = nextIndex[currentIndex]; nextIndex[currentIndex] = nextAvailable; nextAvailable = currentIndex; return value; } } |
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