題目：680. 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.

第一次：
想法是當(dāng)?shù)谝淮斡龅讲煌瑒t比較左下一位是否與右當(dāng)前一致，一致則左＋1；否則比較右＋1位和左當(dāng)前；都不一致則返回false。
458 / 460 test cases passed.
失敗于"cuppucu" 這樣的例子，刪左刪右結(jié)果不同

class Solution {
    public boolean validPalindrome(String s) {
        boolean delete = false;
        int len = s.length();
        if(len <= 2 && len >= 0) return true;
        
        for(int i =0, j = len-1;i<j;i++,j--){
            if(s.charAt(i) != s.charAt(j)){
                if(!delete){
                    delete = true;
                    if(s.charAt(i+1) == s.charAt(j)){
                        i++;
                    }else if(s.charAt(i) == s.charAt(j-1)){
                        j--;
                    }else return false;
                }else{
                    return false;
                }
            }
        }
        return true;
    }
}

第二次嘗試：
460 / 460 test cases passed.
Runtime: 349 ms
效率非常低下

class Solution {
    public boolean validPalindrome(String s) {
        boolean delete = false;
        int len = s.length();
        if(len <= 2 && len >= 0) return true;
                   

        for(int i =0, j = len-1;i<j;i++,j--){
            if(s.charAt(i) != s.charAt(j)){
                    if(s.charAt(i+1) == s.charAt(j)){ //左刪一位可以的情況下，先嘗試這種辦法
                        int i2 = i+1, j2 = j;
                        while(i2 < j2){
                            if(s.charAt(i2) != s.charAt(j2)) break;
                            i2++;j2--;
                            if(i2>=j2) return true;
                        }
                    }else{//左刪不可以，或者左刪嘗試失敗時(shí) 嘗試右刪,右刪再失敗就失敗
                        j = j-1;
                        while(i < j){
                            if(s.charAt(i) != s.charAt(j)) return false;
                            i++;j--;
                        }
                    }
            }
        }
        return true;      
    }
}

分享一下優(yōu)秀版：
所以這么簡單的道理，到底為什么自己想的那么復(fù)雜呢？

class Solution {
    public boolean validPalindrome(String s) {
        int l = -1, r = s.length();
        while (++l < --r) {
            if (s.charAt(l) != s.charAt(r)) return isPalindromic(s, l, r+1) || isPalindromic(s, l-1, r);//左刪or右刪
        }
        return true;
    }
    private boolean isPalindromic(String s, int l, int r) {
        while (++l < --r) {
            if (s.charAt(l) != s.charAt(r)) return false;
        }
        return true;
    }
}