Q5 Longest Palindromic Substring
Given a string s, return the longest palindromic substring in s.
Example 1:
Input: s = "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Example 2:
Input: s = "cbbd"
Output: "bb"
Example 3:
Input: s = "a"
Output: "a"
Example 4:
Input: s = "ac"
Output: "a"
Constraints:
1 <= s.length <= 1000sconsist of only digits and English letters (lower-case and/or upper-case),
解法及注释
class Solution {
public String longestPalindrome(String s) {
//pre-check
if(s == null || s.length() == 0)
return "";
int start = 0, end = 0;
for(int i = 0; i < s.length(); i++) {
int oddLen = getRes(s, i, i);
int evenLen = getRes(s, i, i + 1);
int len = Math.max(oddLen, evenLen);
if(len > end - start) {
start = i - (len - 1) / 2;
end = i + len / 2;
}
}
return s.substring(start, end + 1);
}
private int getRes(String str, int left, int right) {
int l = left, r = right;
while(l >= 0 && r < str.length() && str.charAt(l) == str.charAt(r)) {
l--;
r++;
}
return r - l - 1;
}
}
Q516 Longest Palindromic Subsequence
Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.
Example 1:
Input:
"bbbab"
Output:
4
One possible longest palindromic subsequence is "bbbb".
Example 2:
Input:
"cbbd"
Output:
2
One possible longest palindromic subsequence is "bb".
Constraints:
1 <= s.length <= 1000sconsists only of lowercase English letters.
解法及注释
class Solution {
public int longestPalindromeSubseq(String s) {
//pre-check
if(s == null || s.length() == 0)
return 0;
//DP solution
int n = s.length();
//states
int[][] f = new int[n][n];
for(int len = 1; len <= n; len++) {
for(int i = 0; i + len - 1 < n; i++) {
int j = i + len - 1;
if(len == 1)
f[i][j] = 1;
else if(s.charAt(i) == s.charAt(j))
f[i][j] = f[i+1][j-1] + 2;
else
f[i][j] = Math.max(f[i+1][j], f[i][j-1]);
}
}
return f[0][n-1];
}
}
Q647 Palindromic Substrings
Given a string, your task is to count how many palindromic substrings in this string.
The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
Example 1:
Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".
Example 2:
Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
Intuition
Let N be the length of the string. The middle of the palindrome could be in one of 2N - 1 positions: either at letter or between two letters.
For each center, let's count all the palindromes that have this center. Notice that if [a, b] is a palindromic interval (meaning S[a], S[a+1], ..., S[b] is a palindrome), then [a+1, b-1] is one too.
Algorithm
For each possible palindrome center, let's expand our candidate palindrome on the interval [left, right] as long as we can. The condition for expanding is left >= 0 and right < N and S[left] == S[right]. That means we want to count a new palindrome S[left], S[left+1], ..., S[right].
解法及注释
class Solution {
public int countSubstrings(String s) {
//pre-check
if(s == null || s.length() == 0)
return 0;
//start with center
int n = s.length(), ans = 0;
for(int center = 0 ; center <= 2 * n - 1; ++center) {
int left = center / 2;
int right = left + center % 2;
while(left >= 0 && right < n && s.charAt(left) == s.charAt(right)) {
left--;
right++;
ans++;
}
}
return ans;
}
}
Q25 Valid Palindrome
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
Note: For the purpose of this problem, we define empty string as valid palindrome.
Example 1:
Input: "A man, a plan, a canal: Panama"
Output: true
Example 2:
Input: "race a car"
Output: false
Constraints:
sconsists only of printable ASCII characters.
解法及注释
class Solution {
public boolean isPalindrome(String s) {
//pre-check
if(s == null || s.length() == 0)
return true;
int i = 0, j = s.length() - 1;
while(i <= j) {
if(!isAlphanumeric(s.charAt(i)))
i++;
else if (!isAlphanumeric(s.charAt(j)))
j--;
else {
if(Character.toUpperCase(s.charAt(i)) == Character.toUpperCase(s.charAt(j))) {
i++;
j--;
} else
return false;
}
}
return true;
}
private boolean isAlphanumeric(char c) {
if(c >= 97 && c <= 122)
return true;
else if( c >= 65 && c <= 90)
return true;
else if(c >= 48 && c <= 57)
return true;
return false;
}
}
Q680 Valid Palindrome II
Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
Example 1:
Input: "aba"
Output: True
Example 2:
Input: "abca"
Output: True
Explanation: You could delete the character 'c'.
Note:
- The string will only contain lowercase characters a-z. The maximum length of the string is 50000.
解法及注释
class Solution { public boolean validPalindrome(String s) { //pre-check if(s == null || s.length() == 0 || s.length() > 10e5) { return false; } int left = 0, right = s.length() - 1; while(left < right) { if(s.charAt(left) != s.charAt(right)) { return validRestString(s, left + 1, right) || validRestString(s, left, right - 1); } left++; right--; } return true; } private boolean validRestString(String s, int left, int right) { String str = s.substring(left, right + 1); String reverse = new StringBuilder(str).reverse().toString(); return str.equals(reverse); }}
you may also use two pointers as below
class Solution { public boolean validPalindrome(String s) { //pre-check if(s == null || s.length() == 0 || s.length() > 10e5) { return false; } int left = 0, right = s.length() - 1; while(left <= right) { if(s.charAt(left) == s.charAt(right)) { left++; right--; } else { return isPalindrome(s, left + 1, right) || isPalindrome(s, left, right - 1); } } return true; } private boolean isPalindrome(String str, int left, int right) { while(left <= right) { if(str.charAt(left) != str.charAt(right)) { return false; } left++; right--; } return true; }}
