Day1:剑指 Offer 30. 包含min函数的栈
作者:互联网
题目链接
题目:定义栈的数据结构,请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中,调用 min、push 及 pop 的时间复杂度都是 O(1)。
class MinStack {
//stack1负责添加元素,stack2负责维护当前最小的元素
Deque<Integer> stack1;
Deque<Integer> stack2;
/** initialize your data structure here. */
public MinStack() {
stack1 = new LinkedList<>();
stack2 = new LinkedList<>();
}
public void push(int x) {
stack1.push(x);
if(stack2.isEmpty()){
stack2.push(x);
}else{
int temp = stack2.peek();
stack2.push(temp > x ? x : temp);
}
}
public void pop() {
stack1.pop();
stack2.pop();
}
public int top() {
return stack1.peek();
}
public int min() {
return stack2.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.min();
*/
标签:Offer,int,30,pop,Day1,push,stack2,public,stack1 来源: https://blog.csdn.net/upset_poor/article/details/121768981