数据结构之SWUSTOJ962: 括号匹配问题

题目：

思路：

代码：

#include
#include
#include
#include
typedef struct Stack
{
	char* data;
	int top;//栈顶元素坐标
	int capacity;//栈的容量
}ST;
void STInit(ST* ps)
{
	ps->data = NULL;
	ps->top = 0;//ps->top=-1;
	//初始化时top给0，意味着top指向栈顶数据的下一个
	//初始化时top给1，意味着top指向栈顶数据
	ps->capacity = 0;//初始化容量给0
}
void StackPush(ST* ps, char x)//压栈/入栈 *** 作
{
	if (ps->capacity == ps->top)
	{
		int newCapacity = ps->capacity == 0 ? 4 : ps->capacity * 2;//三目 *** 作符
		//如果ps->capacity等于0，那么就给4个字节的空间，如果不等于0就扩容两倍
		char* tmp = (char*)realloc(ps->data, sizeof(char)*newCapacity);
		if (tmp == NULL)//扩容失败退出程序
		{
			exit(-1);
		}
		ps->capacity = newCapacity;//把capacity换成新的
		ps->data = tmp;//data也指向新的空间
	}
	ps->data[ps->top] = x;
	ps->top++;
}
void StackPop(ST* ps)
{
	ps->top--;
}
char StackTop(ST* ps)
{
	return ps->data[ps->top - 1];
}
bool StackEmpty(ST* ps)
{
	return ps->top == 0;//如果top等于0，返回true
}
int main()
{
	char arr[100] = { 0 };
	scanf("%s", arr);
	getchar();
	int n = strlen(arr);
	ST List;
	STInit(&List);
	int flag = 0;
	for (int i = 0; i < n; i++)
	{
		if (arr[i] == '[' || arr[i] == '(')
		{
			StackPush(&List, arr[i]);
		}
		else
		{
			if (StackEmpty(&List))
			{
				flag = 1;//为空时置为1
				break;
			}
			else if (arr[i] == ']' && StackTop(&List) == '[')
			{
				StackPop(&List);
			}
			else if (arr[i] == ')' && StackTop(&List) == '(')
			{
				StackPop(&List);
			}
			else
			{
				flag = 1;//当右括号和左括号不匹配时置为1
				break;
			}
		}
	}
	if (flag == 1 || List.top != 0)//栈不为空栈时或者flag=1时返回no
		printf("NO");
	else
		printf("YES");

	return 0;
}