标签：char col run convert rac mat ++ art seq
Given a string which we can delete at most k, return whether you can make a palindrome.
For example, given ‘waterrfetawx‘ and a k of 2, you could delete f and x to get ‘waterretaw‘.
Notice the recursive structure of this problem:
This takes O(2min(n, k)) time, where n is the length of the original string. This is because we call k_palindrome twice on each subproblem, and on each call, either k or the string gets reduced by at least 1.
class Solution { public boolean canMakePalindrome(String s, int k) { return canMakePalindromeHelper(s, 0, s.length() - 1, k); } private boolean canMakePalindromeHelper(String s, int start, int end, int k) { while(start < end) { if(s.charAt(start) == s.charAt(end)) { start++; end--; } else { break; } } if(start == end) { return true; } if(k == 0) { return false; } return canMakePalindromeHelper(s, start + 1, end, k - 1) || canMakePalindromeHelper(s, start, end - 1, k - 1); } }
Dynamic Programming, O(n^2 * k) runtime, O(n^2 * k) space
dp[i][j][k]: if the substring s[i - j] can make a palindrome with at most k deletions.
dp[i][j][k] = dp[i + 1][j - 1][k] if s[i] == s[j]; dp[i][j][k] = dp[i + 1][j][k - 1] || dp[i][j - 1][k - 1] if s[i] != s[j];
substring of length 1 and 2 needs to be initialized first.
class Solution { public boolean canMakePalindrome(String s, int k) { int n = s.length(); boolean[][][] dp = new boolean[n][n][k + 1]; for(int i = 0; i < n; i++) { for(int del = 0; del <= k; del++) { dp[i][i][del] = true; } } for(int startIdx = 0; startIdx <= n - 2; startIdx++) { int endIdx = startIdx + 1; for(int del = 0; del <= k; del++) { if(del == 0 && s.charAt(startIdx) == s.charAt(endIdx) || del > 0) { dp[startIdx][endIdx][del] = true; } } } for(int del = 0; del <= k; del++) { for(int l = 3; l <= n; l++) { for(int startIdx = 0; startIdx <= n - l; startIdx++) { int endIdx = startIdx + l - 1; if(startIdx == 1 && endIdx == 9 && del == 1) { System.out.println(); } if(s.charAt(startIdx) == s.charAt(endIdx)) { dp[startIdx][endIdx][del] = dp[startIdx + 1][endIdx - 1][del]; } else if(del > 0) { dp[startIdx][endIdx][del] = (dp[startIdx + 1][endIdx][del - 1] || dp[startIdx][endIdx - 1][del - 1]); } } } } for(int del = 0; del <= k; del++) { if(dp[0][n - 1][del]) { return true; } } return false; } }
Dynamic Programming, convert to the Longest Palindromic Subsequence problem. O(n^2) runtime, O(n^2) space
Find the length of the longest palindromic subsequence using the same optimal substructure, let it be longest. If s.length() - longest <= k, then return true, otherwise return false.
class Solution { public boolean canMakePalindrome(String s, int k) { int longest = longestPalindromeSubseq(s); return s.length() - longest <= k; } private int longestPalindromeSubseq(String s) { if(s == null || s.length() == 0){ return 0; } int n = s.length(); int[][] dp = new int[n][n]; for(int i = 0; i < n; i++){ dp[i][i] = 1; } for(int i = 0; i < n - 1; i++){ dp[i][i + 1] = s.charAt(i) == s.charAt(i + 1) ? 2 : 1; } for(int l = 3; l <= n; l++){ for(int startIdx = 0; startIdx <= n - l; startIdx++){ int endIdx = startIdx + l - 1; if(s.charAt(startIdx) == s.charAt(endIdx)){ dp[startIdx][endIdx] = dp[startIdx + 1][endIdx - 1] + 2; } else{ dp[startIdx][endIdx] = Math.max(dp[startIdx + 1][endIdx], dp[startIdx][endIdx - 1]); } } } return dp[0][n - 1]; } }
Palindrome Algorithm Questions
标签：char col run convert rac mat ++ art seq
原文地址：https://www.cnblogs.com/lz87/p/10619175.html
