680. Valid Palindrome II

最新推荐文章于 2025-05-28 18:06:51 发布

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最新推荐文章于 2025-05-28 18:06:51 发布

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680.Valid Palindrome II

Easy

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Given a string s, return true if the s can be palindrome after deleting at most one character from it.

Example 1:

Input: s = "aba"
Output: true

Example 2:

Input: s = "abca"
Output: true
Explanation: You could delete the character 'c'.

Example 3:

Input: s = "abc"
Output: false

这是一个双指针问题，但是要注意删除后面或者前面的数有两个选择，会出现不该删除该字母但是结果旁边，例如，我下面这段代码就是这样的问题：

class Solution {
    public boolean validPalindrome(String s) {
        int high = s.length()-1, low = 0;
        boolean t = true, isPalindrome = true;
        while (low < high){
            if (s.charAt(low) == s.charAt(high)) {low++;high--;continue;}
            else if (t && (s.charAt(low) == s.charAt(high-1))) {low ++; high = high-2; t = false;continue;}
            else if (t && (s.charAt(low+1) == s.charAt(high))) {low = low+2; high--; t = false;continue;}
            
            else {isPalindrome = false; break;}
        }
        return isPalindrome;
    }
}

改过来之后，再定义一个函数来判断

class 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 static boolean isPalindrome(String s, int i,int j){
        while(i<j){
            if (s.charAt(i++)!=s.charAt(j--)) return false;
        }
        return true;
    }
    
}

大公搞成了！！