zoukankan      html  css  js  c++  java
  • (微软100题)2.设计包含min 函数的栈。

    #include <iostream>
    using namespace std;
    /*2.设计包含min 函数的栈。
    定义栈的数据结构,要求添加一个min 函数,能够得到栈的最小元素。
    要求函数min、push 以及pop 的时间复杂度都是O(1)。
    ANSWER:
    Stack is a LIFO data structure. When some element is popped from the stack, the status will recover to the original status as before that element was pushed. So we can recover the minimum element, too. 
    */
    struct MinStackElement
    {
    	int data;
    	int min;
    };
    
    struct MinStack 
    {
    	MinStackElement * data;
    	int size;
    	int top;
    };
    
    MinStack MinStackInit(int maxSize) 
    {
    	MinStack stack;
    	stack.size = maxSize;
    	stack.data = (MinStackElement*) malloc(sizeof(MinStackElement)*maxSize);
    	stack.top = 0;
    	return stack;
    }
    void MinStackFree(MinStack stack) 
    {
    	free(stack.data);
    }
    void MinStackPush(MinStack stack, int d) 
    {
    	if (stack.top == stack.size) 
    		cout<<"out of stack space.";
    	MinStackElement* p = &stack.data[stack.top];
    	p->data = d;
    	p->min = (stack.top==0?d : stack.data[stack.top-1].min);
    	if (p->min > d) p->min = d;
    	stack.top ++;
    }
    int MinStackPop(MinStack stack) {
    	if (stack.top == 0) 
    		cout<<"stack is empty!";
    	return stack.data[--stack.top].data;
    }
    int MinStackMin(MinStack stack) {
    	if (stack.top == 0) 
    		cout<<"stack is empty!";
    	return stack.data[stack.top-1].min;
    }
    
    void main()
    {
    }
    

  • 相关阅读:
    关于数组的算法-编程之美读后感-1
    java学习笔记之线程1
    java学习笔记之IO一()
    java学习笔记之泛型
    java学习笔记之正则表达式
    Thinking in java学习笔记之String的不可变性
    Thinking in java学习笔记之map的应用
    Thinking in java学习笔记之set
    scrapy之中间件
    Linux之Redis-redis哨兵集群详解
  • 原文地址:https://www.cnblogs.com/byfei/p/6389846.html
Copyright © 2011-2022 走看看