设计一个支持 push,pop,top 操作,并能在常数时间内检索到最小元素的栈。 push(x) -- 将元素 x 推入栈中。 pop() -- 删除栈顶的元素。 top() -- 获取栈顶元素。 getMin() -- 检索栈中的最小元素。 示例: MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> 返回 -3. minStack.pop(); minStack.top(); --> 返回 0. minStack.getMin(); --> 返回 -2. 解法1: class MinStack { Stack<Integer> stack=null; Stack<Integer> minStack=null; /** initialize your data structure here. */ public MinStack() { stack=new Stack<>(); minStack=new Stack<>(); } public void push(int x) { if (stack.isEmpty()&&minStack.isEmpty()){ stack.push(x); minStack.push(x); }else { int y=minStack.peek(); stack.push(x); if (y<x){ minStack.push(y); }else { minStack.push(x); } } } public void pop() { stack.pop(); minStack.pop(); } public int top() { return stack.get(stack.size()-1); } public int getMin() { return minStack.peek(); } } 解法2:class MinStack { Stack<Integer> stack=null; /** initialize your data structure here. */ public MinStack() { stack=new Stack<>(); } public void push(int x) { if (stack.isEmpty()){ stack.push(x); stack.push(x); }else { int y=stack.peek(); stack.push(x); if (y<x){ stack.push(y); }else { stack.push(x); } } } public void pop() { stack.pop(); stack.pop(); } public int top() { return stack.get(stack.size()-2); } public int getMin() { return stack.peek(); } }