LeetCode 1216. Valid Palindrome III

原题链接在这里：https://leetcode.com/problems/valid-palindrome-iii/

题目：

Given a string s and an integer k, find out if the given string is a K-Palindrome or not.

A string is K-Palindrome if it can be transformed into a palindrome by removing at most k characters from it.

Example 1:

Input: s = "abcdeca", k = 2
Output: true
Explanation: Remove 'b' and 'e' characters.

Constraints:

  • 1 <= s.length <= 1000
  • s has only lowercase English letters.
  • 1 <= k <= s.length

题解：

Find the longest palindrome from s.

And check the difference between s and longest palindrome length, if it is <= k, then return true.

Time Complexity: O(n^2). n = s.length().

Space: O(n^2).

AC Java:

 1 class Solution {
 2     public boolean isValidPalindrome(String s, int k) {
 3         if(s == null || s.length() <= k){
 4             return true;
 5         }
 6         
 7         int lp = findLongestPalin(s);
 8         return s.length() - lp <= k;
 9     }
10     
11     private int findLongestPalin(String s){
12         if(s == null || s.length() == 0){
13             return 0;
14         }
15         
16         int n = s.length();
17         int [][] dp = new int[n][n];
18         for(int i = n-1; i>=0; i--){
19             dp[i][i] = 1;
20             for(int j = i+1; j<n; j++){
21                 if(s.charAt(i) == s.charAt(j)){
22                     dp[i][j] = dp[i+1][j-1] + 2;
23                 }else{
24                     dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
25                 }
26             }
27         }
28         
29         return dp[0][n-1];
30     }
31 }

AC Python:

 1 class Solution:
 2     def isValidPalindrome(self, s: str, k: int) -> bool:
 3         if not s:
 4             return True
 5         
 6         longestPalinLen = self.findLongestPalin(s)
 7         return len(s) - longestPalinLen <= k
 8     
 9     def findLongestPalin(self, s):
10         if not s:
11             return 0
12         
13         n = len(s)
14         dp = [[0 for _ in range(n)] for _ in range(n)]
15         for i in range(n - 1, -1, -1):
16             dp[i][i] = 1
17             for j in range(i + 1, n):
18                 if s[i] == s[j]:
19                     dp[i][j] = dp[i + 1][j - 1] + 2
20                 else:
21                     dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
22         return dp[0][n - 1]

类似Longest Palindromic Subsequence, Valid Palindrome, Valid Palindrome II.

posted @ 2019-12-12 11:38  Dylan_Java_NYC  阅读(1530)  评论(0)  编辑  收藏  举报