leetcode 55. Min Stack 如果有新的更小Min，直接在stack里加入一行Min.

梦里梦外; 2022-09-24 12:22 30阅读 0赞

### 55. Min Stack ###

Question Editorial Solution

[My Submissions][]

*  Total Accepted: 82427
 *  Total Submissions: 348648
 *  Difficulty: Easy

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.

public class MinStack {
        Stack<Integer> stk = new Stack<Integer>();
        int min=Integer.MAX_VALUE;
        /** initialize your data structure here. */
        public MinStack() {
        }
        
        public void push(int x) {
            if(x<=min){  //如果出现更小，则更新Min
                stk.push(min); //多存一个最小的。。。。。。。。。。。
                min = x;
            }
            stk.push(x);//不管是否最小都入栈
        }
        
        public void pop() {
            if(stk.peek()==min){
                stk.pop();//如果出最小，则出了和最小相同的那个之后，令最小等于下一个
                min=stk.peek();
                stk.pop();
            }
            else stk.pop();
            if(stk.empty()){
                min=Integer.MAX_VALUE;
            }
        }
        
        public int top() {
            return stk.peek();
        }
        
        public int getMin() {
            return min;
        }
    }
    
    /**
     * 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();
     */

[My Submissions]: https://leetcode.com/problems/min-stack/submissions/