Python 有一个需要注意的点，如果出现了类似")("的情况如果只判断左右括号数量是否相等是会出问题的，因此要判断出否出现过这种情况即top是否小于过-1

express = input()
top = -1
flag = True
for ch in express:
    if ch == '(':
        top = top + 1
    
    elif ch == ')':
        top = top - 1
        if top < -1:
            flag = False

if top == -1 and flag:
    print('YES')

else:
    print('NO')

都是大佬捏

st=[""]*100
top=-1
s=input()
for i in s:
    if i=='(':
        top+=1
        st[top]=i
    elif i==')':
        if top==-1:
            print('NO')
            top=-2
            break
        else:
            top-=1
if top>-1:
    print('NO')
elif top==-1:
    print('YES')

括号匹配用栈实现，遇到左括号直接入栈，遇到右括号判断栈是否为空，为空则直接退出，不为空再出栈

a=input()
top=-1
s=[""]*255
flag = True
for i in a:
    if i == "(" : 
        top += 1
        s[top] = 1
    elif i == ")":
        if top == -1 : 
            flag = False
        s[top] = 0
        top -= 1
if top == -1 and flag :
    print("YES")
else:
    print("NO")

${\large\color{#92a1e8}【栈】表达式括号匹配(stack)}$

一个基础的栈

$\texttt{\color{#e51211}难度：入门}$ $\textit{\small\color{#e51211}easy}$

last update:2023/11/7

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${\large\color{#52a1e8}Python：}$

st=[""]*100
top=0
s=input()
flag=True
for i in range(len(s)):
    if s[i]=='(':
        top+=1
    if s[i]==')':
        top-=1
    if(top<0):
        flag=False
        break
if(flag==True and top==0):
    print("YES")
else:
    print("NO")

st=[""]*100
top=-1
flag=True
s=input()
for i in range(len(s)):
    if s[i]=="(":
        top+=1
        st[top]="("
    elif s[i]==")":
        if top==-1:
            flag=False
            break
        else:
            top-=1
if top>=0:
    flag=False
if flag:
    print('YES')
else:
    print("NO")