题解 | #完成括号匹配#

seq = input()
cnt_left = cnt_right = 0
left_add = right_add = 0
for c in seq:
    if c == '[':
        cnt_left += 1
    else:
        cnt_right += 1
    if cnt_right > cnt_left:
        cnt_left = cnt_right = 0
        left_add += 1

cnt_left = cnt_right = 0
for c in seq[::-1]:
    if c == '[':
        cnt_left += 1
    else:
        cnt_right += 1
    if cnt_right < cnt_left:
        cnt_left = cnt_right = 0
        right_add += 1

print('['*left_add + seq + ']'*right_add)

计算不匹配的左括号数量和右括号数量，输出时补上；

括号匹配的过程中，左右括号地位是对称的，所以正向遍历一次，反向遍历一次即可求出不匹配的左右括号数量