85 lines
1.9 KiB
Java
85 lines
1.9 KiB
Java
//定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。
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//
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//
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//
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// 示例:
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//
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// MinStack minStack = new MinStack();
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//minStack.push(-2);
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//minStack.push(0);
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//minStack.push(-3);
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//minStack.min(); --> 返回 -3.
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//minStack.pop();
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//minStack.top(); --> 返回 0.
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//minStack.min(); --> 返回 -2.
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//
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//
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//
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//
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// 提示：
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//
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//
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// 各函数的调用总次数不超过 20000 次
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//
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//
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//
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//
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// 注意：本题与主站 155 题相同：https://leetcode-cn.com/problems/min-stack/
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// Related Topics 栈 设计
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// 👍 114 👎 0
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package leetcode.editor.cn;
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import java.util.Stack;
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//面试题30:包含min函数的栈
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public class BaoHanMinhanShuDeZhanLcof{
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public static void main(String[] args) {
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//测试代码
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// Solution solution = new BaoHanMinhanShuDeZhanLcof().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class MinStack {
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Stack<Integer> stack;
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Stack<Integer> min;
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public MinStack() {
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stack = new Stack<>();
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min = new Stack<>();
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}
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public void push(int val) {
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if (stack.isEmpty()) {
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min.push(val);
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} else {
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min.push(Math.min(min.peek(), val));
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}
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stack.push(val);
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}
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public void pop() {
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stack.pop();
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min.pop();
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}
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public int top() {
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return stack.peek();
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}
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public int min() {
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return min.peek();
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}
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}
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/**
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* Your MinStack object will be instantiated and called as such:
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* MinStack obj = new MinStack();
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* obj.push(x);
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* obj.pop();
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* int param_3 = obj.top();
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* int param_4 = obj.min();
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*/
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//leetcode submit region end(Prohibit modification and deletion)
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} |