LeetCode 155. Min Stack

原题链接在这里: https://leetcode.com/problems/min-stack/

题目： 

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.

题解：

可以用另一个stack, min_stk来维护最小值. push()时检测x是否小于等于min_stk 的peek(), 若true，则同时push进min_stk.

pop()时若该值等于min_stk.peek()则同时对min_stk进行pop()操作。

getMin()返回min_stk.peek()即可。

Note:使用peek(), pop()之前，都要检测stack.isEmpty(), 否则会throw exception.

Time Complexity: getMin(), O(1); push(), O(1); pop(), O(1); top(), O(1).

Space: O(n).

AC Java:

public class MinStack {

    Stack<Integer> stk;
    Stack<Integer> minStk;
    public MinStack() {
        stk = new Stack<Integer>();
        minStk = new Stack<Integer>();
    }
    
    public void push(int x) {
        stk.push(x);
        if(minStk.isEmpty() || minStk.peek()>=x){
            minStk.push(x);
        }
    }
    
    public void pop() {
        if(!stk.isEmpty()){
            int x = stk.pop();
            if(!minStk.isEmpty() && x == minStk.peek()){
                minStk.pop();
            }
        }
    }
    
    public int top() {
        if(!stk.isEmpty()){
            return stk.peek();
        }
        throw new IllegalStateException("Stack is empty!");
    }
    
    public int getMin() {
        if(!minStk.isEmpty()){
            return minStk.peek();
        }
        throw new IllegalStateException("Stack is empty!");
    }
}

/**
 * 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();
 */

类似Max Stack.