【字符串】B042_LC_破壞迴文串（思維）

一、Problem

Given a palindromic string palindrome, replace exactly one character by any lowercase English letter so that the string becomes the lexicographically smallest possible string that isn’t a palindrome.

After doing so, return the final string. If there is no way to do so, return the empty string.

Example 1
Input: palindrome = "abccba"
Output: "aaccba"

Example 2
Input: palindrome = "a"
Output: ""

二、Solution

方法一：思維題

• 找到第一個非 'a' 字符，將其提換成 'b'；如果找不到，則證明字符串全是 'a' ，於是將最後一個字符替換成 'b'，可滿足替換後的字符串字典序最小。
• 給定的字符串是迴文的，所以只需要枚舉 [0, n/2]

class Solution {
    public String breakPalindrome(String palindrome ) {
        char[] s = palindrome.toCharArray();
        int n = s.length;
        if (n == 1)
            return "";
        for (int i = 0; i < n/2; i++) {
            if (s[i] != 'a') {
                s[i] = 'a';
                return new String(s);
            }
        }
        s[n-1] = 'b';
        return new String(s);
    }
}

複雜度分析

• 時間複雜度：$O(n)$，
• 空間複雜度：$O(1)$，