E
1520797111
tags: Stack, Design
双Stack：一个正常stack，另一个minStack存当下level最小值. 注意维护minStack的变化
另外. 如果要maxStack，也是类似做法
```
/*
Implement a stack with min() function, which will return the smallest number in the stack.
It should support push, pop and min operation all in O(1) cost.
Example
push(1)
pop() // return 1
push(2)
push(3)
min() // return 2
push(1)
min() // return 1
Note
min operation will never be called if there is no number in the stack.
Tags Expand
Stack
Thoughts:
using 2 stacks: one regular, the other one trackes min element
MinStack (0 ~ i): for i elements in regular stack, at each ith, the min element is stored at MinStack(i). This means, there can be duplicated mins for different ith.
Note: remember to check if minStack isEmpty(), empty stack does not have peek()
*/
public class MinStack {
private Stack stack;
private Stack minStack;
public MinStack() {
stack = new Stack();
minStack = new Stack();
}
public void push(int number) {
stack.push(number);
if (minStack.isEmpty()) {
minStack.push(number);
} else {
minStack.push(minStack.peek() >= number ? number : minStack.peek());
}
}
public int pop() {
minStack.pop();
return stack.pop();
}
public int min() {
return minStack.peek();
}
}
/*
Regular stack and minStack.
MinStack keeps the minnimum value at current time (there can be duplicates if min value does not change)
*/
class MinStack {
Stack stack;
Stack minStack;
/** initialize your data structure here. */
public MinStack() {
stack = new Stack();
minStack = new Stack();
}
public void push(int x) {
if (minStack.isEmpty() || getMin() > x) {
minStack.push(x);
} else {
minStack.push(getMin());
}
stack.push(x);
}
public void pop() {
if (!stack.isEmpty()) {
stack.pop();
minStack.pop();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.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.getMin();
*/
```