【LeetCode】680. Valid Palindrome II
Difficulty:easy
More:【目录】LeetCode Java实现
Description
https://leetcode.com/problems/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'.
Intuition
1.Make use of two pointers.
Solution
public boolean validPalindrome(String s) {
for(int i = 0, j = s.length()-1; i < j; i++,j--){
if(s.charAt(i) != s.charAt(j))
return isPalindrome(s, i, j-1) || isPalindrome(s, i+1, j);
}
return true;
}
private boolean isPalindrome(String s, int i, int j){
while(i < j){
if(s.charAt(i++) != s.charAt(j--))
return false;
}
return true;
}
Complexity
Time complexity : O(n)
Space complexity : O(1)
What I've learned
1. It's important to learn and make use of the method isPalindrome()
More:【目录】LeetCode Java实现

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