155. Min Stack

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.
 这道题看上去简单,但是注意在出栈的时候,如果min的出去了,怎么破?

class MinStack {
    Stack<Integer> s = new Stack<Integer>();
    Stack<Integer> ms = new Stack<Integer>();
    int min = Integer.MAX_VALUE;
    /** initialize your data structure here. */
    public MinStack() {
        
    }
    
    public void push(int x) {
        s.push(x);
        if(min>x){
            min = x;
        }
        ms.push(min);
    }
    
    public void pop() {
        s.pop();
        ms.pop();
        if(s.isEmpty()){
            min = Integer.MAX_VALUE;
        }
        else{
            min = ms.peek();
        }
        
        
    }
    
    public int top() {
        return s.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();
 */

喜欢的话订阅一个呗~第一时间收到文章更新哟~

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