680. Valid Palindrome II【easy】

680. Valid Palindrome II【easy】

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:

1. The string will only contain lowercase characters a-z. The maximum length of the string is 50000.

错误解法：

 class Solution {
 public:
     bool validPalindrome(string s) {
         int start = ;
         int end = s.length() - ;
         int flag = false;

         while (start <= end) {
             if (s[start] == s[end]) {
                 ++start;
                 --end;
             }
             else {
                 if (!flag) {
                     flag = true;

                     if (s[start + ] == s[end]) {
                         ++start;
                     } else if (s[start] == s[end - ]) {
                         --end;
                     } else {
                         return false;
                     }
                 }
                 else {
                     return false;
                 }
             }
         }

         return true;

     }
 };

没有通过以下测试用例：

"aguokepatgbnvfqmgmlcupuufxoohdfpgjdmysgvhmvffcnqxjjxqncffvmhvgsymdjgpfdhooxfuupuculmgmqfvnbgtapekouga"

考察代码逻辑，发现一开始我们命中了

s[start + 1] == s[end]

这个逻辑，"aguokepatgbnvfqmgmlcupuufxoohdfpgjdmysgvhmvffcnqxjjxqncffvmhvgsymdjgpfdhooxfuupuculmgmqfvnbgtapekouga"，并且设置了标志位，这就坑了，下次发现不相等的情况，就直接返回false了。

其实我们本意是要命中下面的逻辑

s[start] == s[end - 1]

"aguokepatgbnvfqmgmlcupuufxoohdfpgjdmysgvhmvffcnqxjjxqncffvmhvgsymdjgpfdhooxfuupuculmgmqfvnbgtapekouga"，并且设置了标志位，这样就可以满足测试用例。

就是说每次我们用

s[start + 1] == s[end]

和

s[start] == s[end - 1]

其实是一种“或”的关系，所以我们要一判断就判断到底，比如下面的解法一。

解法一：

 class Solution {
     public boolean validPalindrome(String s)
     {
         int left = , right = s.length() - ;

         while (left < right)
         {
             if (s.charAt(left) == s.charAt(right))
             {
                 left++; right--;
             }
             else
             {
                 //remove right
                 int templeft = left, tempright = right - ;

                 while (templeft < tempright)
                 {
                     if (s.charAt(templeft) != s.charAt(tempright)) break;
                     templeft++; tempright--;

                     if (templeft >= tempright) return true;
                 }

                 //remove left
                 left++;

                 while (left < right)
                 {
                     if (s.charAt(left) != s.charAt(right)) return false;
                     left++; right--;
                 }
             }
         }
         return true;
     }
 }

参考@yashar 的代码。

解法二：

 class Solution {
     public boolean validPalindrome(String s)
     {
         return validPalindrome(s, , s.length() - , false);
     }

     private boolean validPalindrome(String s, int left, int right, Boolean mismatch)
     {
         if (right < left) return true;

         if (s.charAt(left) != s.charAt(right))
         {
             if (mismatch == true)
                 return false;

             return validPalindrome(s, left, right - , true) || validPalindrome(s, left + , right, true);
         }
         else
         {
             return validPalindrome(s, left + , right - , mismatch);
         }
     }
 }

递归，参考@yashar 的代码。

解法三：

 class Solution {
 public:
     bool validPalindrome(string s) {
         return valid(s, , s.length() - , );
     }

 private:
     bool valid(string& s, int i, int j, int d) { // d: num of chars you can delete at most
         if (i >= j) return true;
         if (s[i] == s[j])
             return valid(s, i + , j - , d);
         else
             return d >  && (valid(s, i + , j, d - ) || valid(s, i, j - , d - ));
     }
 };

递归，参考@alexander 的代码。