改良stack的push操作，使得从stack中获取最大元素的时间复杂度为O(1) [其实问题挺无聊]

思想：

       在利用push建栈的过程中，边 “建栈” 边 "利用数组保留当前压栈后的最大的那个元素，即只需要比较数组中的前一个元素即可，若比前一个元素大，则保留该元素到数组中，否则保留数组中的前一个元素相同的值。"
       则当前栈中的最大的元素即使该数组中的最后一个元素。时间复杂度为O(1)

解决：g++ 编译环境

/** main.cpp** Created on: 2009-6-18* Author: NeeSky*/#define MAXSIZE 100#include <iostream>using namespace std;/*** The definition of Stack*/struct Stack{ int size; int data[MAXSIZE];};/*** The Global Variables*/Stack stackProblem;int arrayAssistant[MAXSIZE];/*** Push the element to stackProblem ,* and record the element max to arrayAssistant* */inline void push(int element ){ if (stackProblem.size >= MAXSIZE) /*Overflow Obeverse*/ { cerr << "Push:The stackProblem is full ! " << endl; return; } stackProblem.data[stackProblem.size] = element; arrayAssistant[stackProblem.size++] = element > arrayAssistant[stackProblem.size - 1] ? element : arrayAssistant[stackProblem.size - 1]; return;}/*** Pop the element from stackProblem return the element* */inline int pop(){ if (stackProblem.size == 0) { cerr << "Pop: The stackProblem is empty ! " << endl; return -1; } return stackProblem.data[--stackProblem.size];}/*** Fill the array to stack* */void fill_Stack(){ int A[10] = { 0 }; for (int i = 0; i < 10; ++i) { A[i] = rand() ; cout<<A[i]<<" "; } cout<<endl; for (int i = 0; i < 10; i++) { push(A[i]); } return;}/*** O(1) Time Complex: get the max value from Stack* */int getMax_fromStack(){ return arrayAssistant[stackProblem.size - 1];}int main(void ){ fill_Stack(); cout << getMax_fromStack() << endl; return 0;}

 

输出：

41 18467 6334 26500 19169 15724 11478 29358 26962 24464
29358