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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思路:可以用stack自己的函数实现push,pop,top这三个,关键就是getmin不好实现,那么需要另外一个minstack来存储小的数字。
public class MinStack { /** initialize your data structure here. */ private Stack<Integer>stack; private Stack<Integer>minStack; public MinStack() { stack=new Stack<>(); minStack=new Stack<>(); } public void push(int x) {//入栈 stack.push(x); if(minStack.isEmpty()||x<=minStack.peek())//如果为空的或者是入栈的元素不大于栈顶元素均可入minstack。 { minStack.push(x); } } public void pop() { if(stack.peek().equals(minStack.peek())) { minStack.pop();//保证如果某个元素要出栈的时候,刚好在minstack里面,那么出栈。 } stack.pop(); } public int top() { return stack.peek(); } public int getMin() { return minStack.peek(); //因为minstack的栈顶元素永远都是当前栈中所有元素中的最小值,那么直接出栈顶元素即可。 } }