[LeetCode] 2330. Valid Palindrome IV

You are given a 0-indexed string s consisting of only lowercase English letters. In one operation, you can change any character of s to any other character.

Return true if you can make s a palindrome after performing exactly one or two operations, or return false otherwise.

Example 1:

Input: s = "abcdba"
Output: true
Explanation: One way to make s a palindrome using 1 operation is:
- Change s[2] to 'd'. Now, s = "abddba".
One operation could be performed to make s a palindrome so return true.

Example 2:

Input: s = "aa"
Output: true
Explanation: One way to make s a palindrome using 2 operations is:
- Change s[0] to 'b'. Now, s = "ba".
- Change s[1] to 'b'. Now, s = "bb".
Two operations could be performed to make s a palindrome so return true.

Example 3:

Input: s = "abcdef"
Output: false
Explanation: It is not possible to make s a palindrome using one or two operations so return false.

Constraints:

• 1 <= s.length <= 105
• s consists only of lowercase English letters.

有效的回文 IV。

给你一个下标从 0 开始、仅由小写英文字母组成的字符串 s 。在一步操作中，你可以将 s 中的任一字符更改为其他任何字符。

如果你能在 恰 执行一到两步操作后使 s 变成一个回文，则返回 true ，否则返回 false 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/valid-palindrome-iv
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题意是给一个字符串，你只能修改最多两处，请判断你是否能将这个字符串修改成一个回文串。

思路是双指针从两边往中间逼近，应该算是 easy 题。直接给代码。

时间O(n)

空间O(1)

Java实现

class Solution {
    public boolean makePalindrome(String s) {
        int left = 0;
        int right = s.length() - 1;
        int k = 2;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                k--;
            }
            if (k < 0) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

 

LeetCode 题目总结