1. Topic
To define the data structure of the stack, please implement a min function that can get the smallest element of the stack in this type. In this stack, the time complexity of calling min, push and pop is O(1).
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.min(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.min(); --> 返回 -2.
2. Idea
Key point
1. When the top of stack 1 is popped as the minimum value, the minimum value of the whole stack will also be popped.
Therefore, the minimum value also constitutes a stack 2. When the top of stack 1 is pushed as the minimum value, the minimum value stack 2 is also pushed; when the top of stack 1 is used as the minimum value pop, the minimum value stack 2 is also popped.
First of all, two stacks can be used, one
can be used as a minimum value stack, or one stack can be used for both purposes. . . Each entry and exit brings the minimum value and the top element of the stack, and uses two stack frames at a time.
3. Code
class MinStack {
Stack<Integer> A, B;
public MinStack() {
A = new Stack<>();
B = new Stack<>();
}
public void push(int x) {
A.add(x);
if(B.empty() || B.peek() >= x)
B.add(x);
}
public void pop() {
if(A.pop().equals(B.peek()))
B.pop();
}
public int top() {
return A.peek();
}
public int min() {
return B.peek();
}
}
4. Selected comments and problem-solving ideas
class MinStack {
private Stack<Integer> dataStack; // 数据栈
private Stack<Integer> minStack; // 辅助栈，记录每次有元素进栈后或者出栈后，元素的最小值
/** initialize your data structure here. */
public MinStack() {
// 初始化辅助栈和数据栈
dataStack = new Stack<>();
minStack = new Stack<>();
}
public void push(int x) {
// 如果记录当前数据栈中最小值的辅助栈为空，或者最小值小于 x，则将 x 设置为最小值，即进辅助栈
if(minStack.isEmpty() || minStack.peek() > x){
minStack.push(x);
}else{
// 如果数据栈中当前最小值 < x, 则继续将最小值设置为上次的最小值
minStack.push(minStack.peek());
}
dataStack.push(x);// 数据栈，进栈
}
public void pop() {
minStack.pop();// 辅助栈，栈出栈
dataStack.pop();// 数据栈，出栈
}
public int top() {
return dataStack.peek();
}
public int min() {
return minStack.peek();
}
}