难度中等1363收藏分享切换为英文接收动态反馈
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。
实现 MinStack 类:
MinStack() 初始化堆栈对象。void push(int val) 将元素val推入堆栈。void pop() 删除堆栈顶部的元素。int top() 获取堆栈顶部的元素。int getMin() 获取堆栈中的最小元素。
示例 1:
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.
提示:
-231 <= val <= 231 - 1pop、top 和 getMin 操作总是在 非空栈 上调用push, pop, top, and getMin最多被调用 3 * 104 次


-
- public class MinStack {
-
- private final int INIT_SIZE = 100;
-
- private int[] elements;
- private int size;
- private int min; /* 维护一个最小值 */
- private int minCount;
-
-
- /** initialize your data structure here. */
- public MinStack() {
- elements = new int[INIT_SIZE];
- min = Integer.MAX_VALUE;
- }
-
- public void push(int x) {
- ensureCapacity(); /* 扩容检测 */
- elements[size++] = x;
- /* 维护最小值 */
- if(x < min) {
- min = x;
- minCount = 1;
- } else if(x == min) {
- minCount++;
- }
- }
-
- public void pop() {
- var popNum = elements[--size]; /* 被删除的数 */
- /* 维护最小值 */
- if(popNum == min && --minCount == 0) {
- min = Integer.MAX_VALUE;
- for (int i = 0; i < size; i++) {
- min = Math.min(min, elements[i]);
- }
- minCount = 1;
- }
- }
-
- public int top() {
- return elements[size - 1];
- }
-
- public int getMin() {
- return min; /* 直接返回最小值 */
- }
-
- /**
- * 是否需要扩容
- */
- private void ensureCapacity() {
- if(size >= elements.length - 1) {
- elements = Arrays.copyOf(elements, elements.length + (elements.length >> 1));
- }
- }
-
- }
