題目：包含min函數(shù)的棧

定義棧的數(shù)據(jù)結(jié)構(gòu)，請(qǐng)?jiān)谠擃愋椭袑?shí)現(xiàn)一個(gè)能夠得到棧的最小元素的 min 函數(shù)在該棧中，調(diào)用 min、push 及 pop 的時(shí)間復(fù)雜度都是 O(1)。

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：各函數(shù)的調(diào)用總次數(shù)不超過(guò) 20000 次

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題目：

class MinStack {
public:/** initialize your data structure here. */MinStack() {}void push(int x) {}void pop() {}int top() {}int 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->min();*/

解題：

難點(diǎn)在于時(shí)間復(fù)雜度是O(1)，需要雙容器或雙棧來(lái)實(shí)現(xiàn)，v1是普通的，v2是最后一個(gè)數(shù)永遠(yuǎn)是最小的。解題思路可以看這個(gè)https://leetcode-cn.com/problems/bao-han-minhan-shu-de-zhan-lcof/solution/offer30bao-han-minhan-shu-de-zhan-by-log-a6vx/

class MinStack {
private:vector<int> v1;vector<int> v2;public:/** initialize your data structure here. */MinStack() {v1.clear();v2.clear();}void push(int x) {if(v1.size()==0){v2.emplace_back(x);}else{if(x<v2.back()){v2.emplace_back(x);}else{v2.emplace_back(v2.back());}}v1.emplace_back(x);}void pop() {if(v1.size()==0){return;}v1.pop_back();v2.pop_back();}int top() {return v1.back();}int min() {return v2.back();}
};/*** 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();*/

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