数据结构之SWUSTOJ962: 括号匹配问题 and LeetCode20有效的括号

 题目：

20. 有效的括号 - 力扣（LeetCode） (leetcode-cn.com)https://leetcode-cn.com/problems/valid-parentheses/

SWUST这道题基本都是仿照力扣这题，并且力扣这道题测试用例给的多一些，我们就采用力扣给的用例场景进行图解。

力扣思路：

SWUSTOJ代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdbool.h>
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/入栈操作
{
	if (ps->capacity == ps->top)
	{
		int newCapacity = ps->capacity == 0 ? 4 : ps->capacity * 2;//三目操作符
		//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;
}

LeetCode代码：

#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
#include<stdbool.h>

typedef char STDataType;

typedef struct Stack
{
	STDataType* a;
	int top;
	int capacity;
}ST;
void StackInit(ST* ps)
{
	assert(ps);
	ps->a = NULL;
	ps->top = 0;//ps->top=-1
	//初始化时top给的是0，意味着top指向栈顶数据的下一个
	//初始化时top给的是-1，意味着top指向栈顶数据
	ps->capacity = 0;
}
void StackDestroy(ST* ps)
{
	assert(ps);
	free(ps->a);
	ps->a = NULL;
	ps->capacity = ps->top = 0;
}
void StackPush(ST* ps, STDataType x)
{
	assert(ps);
	if (ps->top == ps->capacity)
	{
		int newCapacity = ps->capacity == 0 ? 4 : ps->capacity * 2;
		STDataType* tmp = realloc(ps->a, sizeof(STDataType)*newCapacity);
		if (tmp == NULL)
		{
			printf("realloc fail\n");
			exit(-1);
		}
		ps->a = tmp;
		ps->capacity = newCapacity;
	}
	ps->a[ps->top] = x;
	ps->top++;
}
void StackPop(ST* ps)
{
	assert(ps);
	assert(ps->top > 0);

	ps->top--;
}
STDataType StackTop(ST* ps)
{
	assert(ps);
	assert(ps->top > 0);//assert(!StackEmpty(ps));
	return ps->a[ps->top - 1];
}
bool StackEmpty(ST* ps)
{
	assert(ps);
	/*if (ps->top == 0)
	{
		return true;
	}
	else
	{
		return false;
	}*/
	return ps->top == 0;
}


bool isValid(char * s)
{
    ST st;
    StackInit(&st);
    while(*s)
    {
        if(*s=='('||*s=='{'||*s=='[')
        {
            StackPush(&st,*s);
            s++;
        }
        else
        {
            //遇到右括号了，但是栈里面没有数据
            //说明前面没有左括号，不匹配，返回false
            if(StackEmpty(&st))
            {
                return false;
            }

            STDataType top=StackTop(&st);
            StackPop(&st);
            if((*s=='}'&&top!='{')
            ||(*s==']'&&top!='[')
            ||(*s==')'&&top!='('))
            {
                StackDestroy(&st);
                return false;
            }
            else
            {
                s++;
            }
        }
    }
    //如果栈不是空，说明栈中还有左括号未出栈，没有匹配上，返回的是fasle
    bool ret=StackEmpty(&st);
    StackDestroy(&st);
    return ret;
}