classSolution {publicString shortestPalindrome(String s) {if(s.length() <= 1)returns; String temp= s + "#"+newStringBuilder(s).reverse().toString();int[] kmp =getTable(temp);//get the maximum palin part in s starts from 0returnnewStringBuilder(s.substring(kmp[kmp.length - 1])).reve...
find all the palindromes in substring s[1:end-1], and all the palindromes in substring s[end] So the problem is quite clear, when we do recursion, two things should be considered: 1. stop condition: when the search goes to the last position in the string 2. for loop or while loop:...
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Output: [true,false,false,true,true] Explanation: queries[0] : substring = "d", is palidrome. queries[1] : substring = "bc", is not palidrome. queries[2] : substring = "abcd", is not palidrome after replacing only 1 character. queries[3] : substring = "abcd", could be changed...