题目：原题链接（困难）
标签：动态规划、字符串
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N 2 ) O(N^2) O(N2) | O ( N 2 ) O(N^2) O(N2) | 324ms (53.85%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def isValidPalindrome(self, s: str, k: int) -> bool:
size = len(s)
# dp[i][j] = 令字符串[i,j]变为回文串需要删除的字符数
dp = [[float("inf")] * size for _ in range(size)]
for i in range(size):
dp[i][i] = 0
for length in range(2, size + 1):
for i in range(size - length + 1):
j = i + length - 1
if s[i] == s[j]:
if length == 2:
dp[i][j] = 0
else:
dp[i][j] = dp[i + 1][j - 1]
else:
dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1
return dp[0][-1] <= k
