Foreword
I encountered an interview question, and the general meaning of the question is as follows:
Use two ordinary stacks to implement a special stack, so that the pop, push, and min functions are all operations with a complexity of O(1). The min function is to obtain the minimum value of the current stack.
Initial Thoughts
1. To implement the min function as (1) operation, the first idea at that time was to calculate the current minimum value in advance, so I thought of using a value to save the minimum element in the current stack, and then perform push and pop operations. Maintain this value. In this way, min and push are both O(1), but pop is not. If the current pop-up is the minimum value, you need to find the minimum value of the current element again, which is not O(1).
2. And the above method does not use another stack, so I thought of storing the sorted elements in a stack, and also maintaining this ordered stack during push and pop operations, as shown in the figure:
But in this case, the min operation is O(1), but because the push and pop operations need to maintain this ordered stack, I can't think of a method that can achieve O(1) complexity.
At that time, I thought that the minimum value information must be cached in another stack, but I don’t know whether it was because I didn’t eat or something, so my thinking froze.
Correct solution
I felt very unhappy when I couldn't solve a problem, so while eating, I started thinking about how to fully understand the characteristics of the stack and effectively cache the minimum value information for use in min operations.
The biggest feature of stack operation is that it can only operate on the top element of the stack. It is not just right to use an auxiliary stack to cache the minimum value of each stack operation. In this way, every time a pop operation is performed, both sides can be popped together; because the top element of the auxiliary stack is the smallest value in the current stack, the push operation only needs to compare the pushed element and the top element of the auxiliary stack. In this way, push, pop, and min are all O(1) operations. As shown in the picture:
The text may not be clear. Here is the code. Below is the PHP implementation, which simulates the stack through an array.
<?php /** * 使用一个辅助栈,O(1)复杂度求出栈中的最小数 * @hack 类中通过数组来模拟堆栈 * * @author laiwenhui */ class strack{ /** * 数据栈,存储栈数据; * * @var array */ private $_arrData = array(); /** * 辅助栈,存储数据组栈中每层的最下值信息; * * @var array */ private $_arrMin = array(); /** * 栈顶所在单元 * * @var int */ private $_top=-1; /** * 出栈 * @return bool|int */ public function pop(){ if ($this->_top === -1){ return false; } array_pop($this->_arrMin); $this->_top--; return array_pop($this->_arrData); } /** * 入栈 * @param int $element * @return bool */ public function push($element){ $element = intval($element); //如果栈为空,直接入栈 if ($this->_top === -1){ array_push($this->_arrData, $element); array_push($this->_arrMin, $element); $this->_top++; return true; } //不为空,判断入栈的值是否比最小栈栈顶小 $min = $this->_arrMin[$this->_top]; //比较求出最小值 $currentMin = $element < $min ? $element : $min; //当前栈中最小值入栈 array_push($this->_arrMin, $currentMin); //数据入栈 array_push($this->_arrData, $element); $this->_top++; return true; } /** * 求当前栈空间的最小值 * @return bool|int */ public function min(){ if ($this->_top === -1){ return false; } return $this->_arrMin[$this->_top]; } }
Use as follows:
$obj = new strack();
$obj->push(12);
$obj->push(56);
$obj->push(23);
$obj->push(89);
$obj->push(4);
var_dump($obj->min());
$obj->pop();
var_dump($obj->min());
$obj->push(8);
var_dump($obj->min());
The output is:
int(4)
int(12)
int(8)
OK, requirements met.
Do you have any other better ways to achieve this? If so, please tell me^_^

php把负数转为正整数的方法:1、使用abs()函数将负数转为正数,使用intval()函数对正数取整,转为正整数,语法“intval(abs($number))”;2、利用“~”位运算符将负数取反加一,语法“~$number + 1”。

实现方法:1、使用“sleep(延迟秒数)”语句,可延迟执行函数若干秒;2、使用“time_nanosleep(延迟秒数,延迟纳秒数)”语句,可延迟执行函数若干秒和纳秒;3、使用“time_sleep_until(time()+7)”语句。

php除以100保留两位小数的方法:1、利用“/”运算符进行除法运算,语法“数值 / 100”;2、使用“number_format(除法结果, 2)”或“sprintf("%.2f",除法结果)”语句进行四舍五入的处理值,并保留两位小数。

php字符串有下标。在PHP中,下标不仅可以应用于数组和对象,还可应用于字符串,利用字符串的下标和中括号“[]”可以访问指定索引位置的字符,并对该字符进行读写,语法“字符串名[下标值]”;字符串的下标值(索引值)只能是整数类型,起始值为0。

判断方法:1、使用“strtotime("年-月-日")”语句将给定的年月日转换为时间戳格式;2、用“date("z",时间戳)+1”语句计算指定时间戳是一年的第几天。date()返回的天数是从0开始计算的,因此真实天数需要在此基础上加1。

在php中,可以使用substr()函数来读取字符串后几个字符,只需要将该函数的第二个参数设置为负值,第三个参数省略即可;语法为“substr(字符串,-n)”,表示读取从字符串结尾处向前数第n个字符开始,直到字符串结尾的全部字符。

方法:1、用“str_replace(" ","其他字符",$str)”语句,可将nbsp符替换为其他字符;2、用“preg_replace("/(\s|\ \;||\xc2\xa0)/","其他字符",$str)”语句。

php判断有没有小数点的方法:1、使用“strpos(数字字符串,'.')”语法,如果返回小数点在字符串中第一次出现的位置,则有小数点;2、使用“strrpos(数字字符串,'.')”语句,如果返回小数点在字符串中最后一次出现的位置,则有。


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