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.

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设计一个栈 有一个getMin的方法 每次获取stack中的最小值 时间复杂度O(1)

设最小值为min 关键在于min的维护 

push的时候简单 只要一直取Math.min就可以了

当pop的时候 如果pop的正好是最小值 怎样找到上一次的最小值呢

比如 输入是 2,3,1

pop掉1后 最小值应该变为2 但是怎样找到2呢

比较常规的做法 再用一个minstack 每次只放入比当前min小的值 相当于是维护了一个最小值的集合 

如果pop掉当前最小值 那么min = stack.pop();

class MinStack {
    int min = Integer.MAX_VALUE;
    Stack<Integer> stack = new Stack<Integer>();
    public void push(int x) {
        // only push the old minimum value when the current 
        // minimum value changes after pushing the new value x
        if(x <= min){          
            stack.push(min);
            min=x;
        }
        stack.push(x);
    }

    public void pop() {
        // if pop operation could result in the changing of the current minimum value, 
        // pop twice and change the current minimum value to the last minimum value.
        if(stack.pop() == min) min=stack.pop();
    }

    public int top() {
        return stack.peek();
    }

    public int getMin() {
        return min;
    }
}

第二种做法不是很好想 只用一个stack

当有新的最小值 先push之前的最小值 这样 pop当前最小值后 可以找到上一次的最小值

class MinStack {
    int min = Integer.MAX_VALUE;
    Stack<Integer> stack = new Stack<Integer>();
    public void push(int x) {
        // only push the old minimum value when the current 
        // minimum value changes after pushing the new value x
        if(x <= min){          
            stack.push(min);
            min=x;
        }
        stack.push(x);
    }

    public void pop() {
        // if pop operation could result in the changing of the current minimum value, 
        // pop twice and change the current minimum value to the last minimum value.
        if(stack.pop() == min) min=stack.pop();
    }

    public int top() {
        return stack.peek();
    }

    public int getMin() {
        return min;
    }
}