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剑指 Offer 30. 包含min函数的栈
题目
剑指 Offer 30. 包含min函数的栈
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)。
示例:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.min(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.min(); --> 返回 -2.
提示:
- 各函数的调用总次数不超过 20000 次
代码
C++:
class MinStack {
private:
stack<int> mstack;
int minnum[20000];
int length;
public:
/** initialize your data structure here. */
MinStack() {
minnum[0]=0;
length=0;
}
void push(int x) {
if(length==0)
{
minnum[length+1]=x;
}
else
{
if(minnum[length]<x)
minnum[length+1]=minnum[length];
else
minnum[length+1]=x;
}
length++;
mstack.push(x);
}
void pop() {
mstack.pop();
length--;
}
int top() {
return mstack.top();
}
int min() {
return minnum[length];
}
};
/**
* 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->min();
*/
Go:
type MinStack struct {
min,data []int
}
/** initialize your data structure here. */
func Constructor() MinStack {
return MinStack{}
}
func (this *MinStack) Push(x int) {
if len(this.data)==0 x<this.min[len(this.min)-1]{
this.min=append(this.min,x)
}else{
this.min=append(this.min,this.min[len(this.min)-1])
}
this.data=append(this.data,x)
}
func (this *MinStack) Pop() {
if len(this.data)!=0{
this.data=this.data[:len(this.data)-1]
this.min=this.min[:len(this.min)-1]
}
}
func (this *MinStack) Top() int {
return this.data[len(this.data)-1]
}
func (this *MinStack) Min() int {
return this.min[len(this.min)-1]
}
/**
* Your MinStack object will be instantiated and called as such:
* obj := Constructor();
* obj.Push(x);
* obj.Pop();
* param_3 := obj.Top();
* param_4 := obj.Min();
*/
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