Min Stack

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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.

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思路 1 ：就是多设置一个 minStack来存储一下最小值， push 和 pop 时候判断一下。

思路2 ：这个具有更小的空间复杂度，设置一个 min 作为标记，在 stack 里面存储和 min 的差值。

class MinStack {    Stack<Integer> s = new Stack<Integer>();    Stack<Integer> minStack = new Stack<Integer>();    public void push(int x) {        s.push(x);        if(minStack.isEmpty() || x <= minStack.peek()){//-----            minStack.push(x);        }    }    public void pop() {        int num = s.pop();        if(num == minStack.peek()){            minStack.pop();        }    }    public int top() {        return s.peek();    }    public int getMin() {        return minStack.peek();    }}

public class MinStack {    long min;    Stack<Long> stack;//有减法，所以用long，将数据越界考虑在内    public MinStack(){        stack=new Stack<>();    }    public void push(int x) {        if (stack.isEmpty()){            stack.push(0L);            min=x;            return;        }        stack.push(x-min);//Could be negative if min value needs to change        if (x<min) min=x;    }    public void pop() {        if (stack.isEmpty()) return;        long pop=stack.pop();        if (pop<0)  min=min-pop;//If negative, increase the min value    }    public int top() {        long top=stack.peek();        if (top>0){            return (int)(top+min);        }else{           return (int)(min);        }    }    public int getMin() {        return (int)min;    }}

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