文章作者:Tyan
博客:noahsnail.com | CSDN |
1. 问题描述
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.
Example:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
2. 求解
主要是模拟写一个最小栈。要注意push时可能会输入null。需要使用双栈实现,一个保存数据,一个保存最小值。由于随着数据出栈,最小值是不断变化的,因此需要一个最小值栈来保存最小值。
class MinStack {
private Stack stack = new Stack<>();
private Stack minStack = new Stack<>();
public void push(int x) {
if(minStack.isEmpty() || x <= minStack.peek()) {
minStack.push(x);
}
stack.push(x);
}
public void pop() {
if(stack.peek().equals(minStack.peek())) {
minStack.pop();
}
stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}