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.
题解1
cpp
struct Node {
int lastmin;
int val;
Node* next;
// 构造函数(可选,但推荐)
Node() : lastmin(0), val(0), next(nullptr) {}
};
class MinStack {
private:
int min_val; // 避免与函数 min 冲突,改名为 min_val
Node* top_node; // 指向头节点(哨兵节点)
public:
MinStack() {
top_node = new Node(); // 创建哨兵头节点
min_val = INT_MAX; // 初始最小值设为最大(更健壮)
}
void push(int val) {
// 如果栈为空(即第一个真实元素)
if (top_node->next == nullptr) {
min_val = val;
}
Node* node = new Node();
node->val = val;
node->lastmin = min_val; // 保存当前最小值(用于 pop 时恢复)
min_val = std::min(min_val, val); // 更新全局最小值
node->next = top_node->next;
top_node->next = node;
}
void pop() {
if (top_node->next == nullptr) return; // 安全检查
Node* node = top_node->next;
min_val = node->lastmin; // 恢复上一个最小值
top_node->next = node->next;
delete node; // 释放内存!
}
int top() {
return top_node->next->val;
}
int getMin() {
return min_val;
}
// 可选:析构函数,防止内存泄漏
~MinStack() {
while (top_node != nullptr) {
Node* temp = top_node;
top_node = top_node->next;
delete temp;
}
}
};
/**
* Your MinStack object will be instantiated and called as such:
* MinStack* obj = new MinStack();
* obj->push(val);
* obj->pop();
* int param_3 = obj->top();
* int param_4 = obj->getMin();
*/
官方题解
cpp
class MinStack {
stack<int> x_stack;
stack<int> min_stack;
public:
MinStack() {
min_stack.push(INT_MAX);
}
void push(int x) {
x_stack.push(x);
min_stack.push(min(min_stack.top(), x));
}
void pop() {
x_stack.pop();
min_stack.pop();
}
int top() {
return x_stack.top();
}
int getMin() {
return min_stack.top();
}
};