목록괄호 (1)
Head vs breakz
[code] - 카카오 괄호변환 python
# 괄호변환 def solution(p): if p =="": return "" elif len(p) % 2 == 0: qq = div_val(p,point_val(p)) return qq def point_val(p): a = 0 count = 1 q = list(p) for i in range(len(q)): if q[i]=="(": count +=2 if count == 1: a = i break elif q[i]==")": count -=2 if count == 1: a = i break return a def div_val(p,a): count = [] q = list(p) u =q[:a+1] v =p[a+1:] if check_val(u) ==True : return p[:a+1]+soluti..
Head/Code
2020. 5. 4. 20:17