题目
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:
输入:
["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]]
输出:
[null,null,null,null,-3,null,0,-2]
解释:
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.
解答
class MinStack {private LinkedList<Integer> stack = new LinkedList<>();private LinkedList<Integer> minStack = new LinkedList<>();/** initialize your data structure here. */public MinStack() {}public void push(int x) {if (stack.isEmpty()) {minStack.addFirst(x);}else {minStack.addFirst(Math.min(minStack.getFirst(), x));}stack.addFirst(x);}public void pop() {stack.removeFirst();minStack.removeFirst();}public int top() {return stack.getFirst();}public int getMin() {return minStack.getFirst();}
}
要点
常数级操作,意味着各个方法均不能对容器进行遍历,唯一想到的好办法即是把最小值保存下来,便于访问。