力扣:面试题 03.02:栈的最小值

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huangge1199 2021-04-01 14:18:53 +08:00
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//请设计一个栈除了常规栈支持的pop与push函数以外还支持min函数该函数返回栈元素中的最小值执行pushpop和min操作的时间复杂度必须为O(
//1) 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. Related Topics
// 👍 41 👎 0
package leetcode.editor.cn;
import java.util.Stack;
//面试题 03.02:栈的最小值
public class MinStackLcci {
public static void main(String[] args) {
//测试代码
// Solution solution = new MinStackLcci().new Solution();
}
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
class MinStack {
class Data {
int num;
int min;
public Data(int num, int min) {
this.num = num;
this.min = min;
}
public int getNum() {
return num;
}
public void setNum(int num) {
this.num = num;
}
public int getMin() {
return min;
}
public void setMin(int min) {
this.min = min;
}
}
Stack<Data> stack;
public MinStack() {
stack = new Stack<>();
}
public void push(int val) {
if (stack.isEmpty()) {
stack.push(new Data(val,val));
} else {
stack.push(new Data(val, Math.min(val, stack.peek().min)));
}
}
public void pop() {
stack.pop();
}
public int top() {
return stack.peek().num;
}
public int getMin() {
return stack.peek().min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
//leetcode submit region end(Prohibit modification and deletion)
}

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<p>请设计一个栈除了常规栈支持的pop与push函数以外还支持min函数该函数返回栈元素中的最小值。执行push、pop和min操作的时间复杂度必须为O(1)。</p><br><p><strong>示例:</strong><pre>MinStack minStack = new MinStack();<br>minStack.push(-2);<br>minStack.push(0);<br>minStack.push(-3);<br>minStack.getMin(); --> 返回 -3.<br>minStack.pop();<br>minStack.top(); --> 返回 0.<br>minStack.getMin(); --> 返回 -2.</pre></p><div><div>Related Topics</div><div><li></li></div></div>\n<div><li>👍 41</li><li>👎 0</li></div>