|
| 1 | +""" |
| 2 | +[문제풀이] |
| 3 | +# Inputs |
| 4 | +string s |
| 5 | +
|
| 6 | +# Outputs |
| 7 | +the number of palindromic substrings |
| 8 | +
|
| 9 | +# Constraints |
| 10 | +1 <= s.length <= 1000 |
| 11 | +s consists of lowercase English letters. |
| 12 | +
|
| 13 | +# Ideas |
| 14 | +부분 문자열 중 팰린드롬 인거 |
| 15 | +순열? |
| 16 | +10^3 => 시초 예상 |
| 17 | +
|
| 18 | +코드를 짜보니 O(n^3) 나오긴하는데 우선 정답 |
| 19 | +
|
| 20 | +[회고] |
| 21 | +
|
| 22 | +""" |
| 23 | + |
| 24 | + |
| 25 | +class Solution: |
| 26 | + def countSubstrings(self, s: str) -> int: |
| 27 | + ret = 0 |
| 28 | + |
| 29 | + for num in range(1, len(s) + 1): |
| 30 | + for i in range(len(s) - num + 1): |
| 31 | + ss = s[i:i + num] |
| 32 | + if ss == ss[::-1]: |
| 33 | + ret += 1 |
| 34 | + |
| 35 | + return ret |
| 36 | + |
| 37 | +# 해설보고 스스로 풀이 |
| 38 | + |
| 39 | +class Solution: |
| 40 | + def countSubstrings(self, s: str) -> int: |
| 41 | + dp = {} |
| 42 | + |
| 43 | + for start in range(len(s)): |
| 44 | + for end in range(start, -1, -1): |
| 45 | + if start == end: |
| 46 | + dp[(start, end)] = True |
| 47 | + |
| 48 | + elif start + 1 == end: |
| 49 | + dp[(start, end)] = s[start] == s[end] |
| 50 | + |
| 51 | + else: |
| 52 | + dp[(start, end)] = dp[(start + 1, end - 1)] and s[start] == s[end] |
| 53 | + |
| 54 | + return dp.values().count(True) |
| 55 | + |
| 56 | +# 기존 값 재활용하려면 end 부터 세야하는게 이해가 안감 |
| 57 | +# -> 다시 풀이 |
| 58 | +class Solution: |
| 59 | + def countSubstrings(self, s: str) -> int: |
| 60 | + dp = {} |
| 61 | + |
| 62 | + for end in range(len(s)): |
| 63 | + for start in range(end, -1, -1): |
| 64 | + if start == end: |
| 65 | + dp[(start, end)] = True |
| 66 | + |
| 67 | + elif start + 1 == end: |
| 68 | + dp[(start, end)] = s[start] == s[end] |
| 69 | + |
| 70 | + else: |
| 71 | + dp[(start, end)] = dp[(start + 1, end - 1)] and s[start] == s[end] |
| 72 | + |
| 73 | + return list(dp.values()).count(True) |
| 74 | + |
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