stack with min operation

最新推荐文章于 2022-06-02 14:12:00 发布

原创最新推荐文章于 2022-06-02 14:12:00 发布 · 172 阅读

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本文介绍了一种特殊的栈结构，该栈除了基本的push和pop操作外，还额外提供了一个min函数用以返回栈内最小元素。这种特殊的数据结构在进行push、pop及min操作时均能保证O(1)的时间复杂度。通过维护每个节点的当前最小值，确保了在任何时候都能快速找到最小元素。

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How would you design a stack which, in addition to push and pop, also has a function
min which returns the minimum element? Push, pop and min should all operate in
O(1) time.



public class StackWithMin extends Stack<NodeWithMin> {
public void push(int value) {
int newMin = Math.min(value, min());
super.push(new NodeWithMin(value, newMin));
}
}
public int min() {

if (this.isEmpty()) {

return Integer.MAX_VALUE;

} else {

return peek().min;

}
}
class NodeWithMin {
public int value;
public int min;
public NodeWithMin(int v, int min){
value = v;
this.min = min;
}
}