剑指offer 面试题21:包含min函数的栈

题目：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)。

#include <stdio.h>
#include <stdlib.h>
#define  N 10
typedef struct
{
	int stack[N];
	int auStack[N];
	int top1, top2;
}Stack;


void push(Stack  *, int);
void pop(Stack *);
int min_num(Stack *);

int main()
{
	
	Stack s;
	Stack *p = &s;
	s.top1 = 0;
	s.top2 = 0;

	
	push(p,4);
	push(p,5);
	push(p,6);
	push(p,3);
	push(p,1);
	push(p,1);
	
	
//	pop(p);
//	pop(p);
//	pop(p);
	printf("%d\n", min_num(p));
	system("pause");
	return 0;
}

void push(Stack *s, int a)
{
	if (s->top1 == N)
	{
		printf("stack is full!\n");
		return;
	}

	s->stack[s->top1++] = a;
	if (s->top2 == 0)
	{
		s->auStack[s->top2++] = a; 
	}
	else
	{
		if (a <= s->auStack[s->top2 - 1])
		{
			s->auStack[s->top2++] = a;
		}
	}
}

void pop(Stack *s)
{
	if (s->top1 == 0)
	{
		printf("stack is empty!\n");
		return;
	}

	if (s->stack[--s->top1] == s->auStack[s->top2 - 1])
	{
		--s->top2;
	}
}

int min_num(Stack *s)
{
	return s->auStack[s->top2 - 1];
}