155. Min Stack

Description

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.

Solution

Two stacks

用一个minStack来辅助记录，当push时与minStack栈顶进行比较，比栈顶值小才入栈，维持minStack中栈顶到栈底是一个递增的关系。pop的时候也需要与minStack栈顶比较。

class MinStack {
    private Stack stack;
    
    private Stack minStack;
    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack();
        minStack = new Stack();
    }
    
    public void push(int x) {
        stack.push(x);
        if (x <= getMin()) {
            minStack.push(x);
        }
    }
    
    public void pop() {
        if (stack.isEmpty()) return;
        
        int val = stack.pop();
        if (val == getMin()) {
            minStack.pop();
        }
    }
    
    public int top() {
        if (stack.isEmpty()) {
            return Integer.MAX_VALUE;
        }
        return stack.peek();
    }
    
    public int getMin() {
        if (minStack.isEmpty()) return Integer.MAX_VALUE;
        return minStack.peek();
    }
}

/**
 * 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();
 */