680. Valid Palindrome IIeasy

Posted 2020-10-08 abc_begin

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：

 1 class Solution {
 2 public:
 3     bool validPalindrome(string s) {
 4         int start = 0;
 5         int end = s.length() - 1;
 6         int flag = false;
 7         
 8         while (start <= end) {
 9             if (s[start] == s[end]) {
10                 ++start;
11                 --end;
12             }
13             else {
14                 if (!flag) {
15                     flag = true;
16                     
17                     if (s[start + 1] == s[end]) {
18                         ++start;
19                     } else if (s[start] == s[end - 1]) {
20                         --end;
21                     } else {
22                         return false;
23                     } 
24                 } 
25                 else {
26                     return false;
27                 }
28             }
29         }
30         
31         return true;
32         
33     }
34 };

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：

 1 class Solution {
 2     public boolean validPalindrome(String s)
 3     {
 4         int left = 0, right = s.length() - 1;
 5 
 6         while (left < right)
 7         {
 8             if (s.charAt(left) == s.charAt(right))
 9             {
10                 left++; right--;
11             }
12             else
13             {
14                 //remove right
15                 int templeft = left, tempright = right - 1;
16 
17                 while (templeft < tempright)
18                 {
19                     if (s.charAt(templeft) != s.charAt(tempright)) break;
20                     templeft++; tempright--;
21 
22                     if (templeft >= tempright) return true;
23                 }
24 
25                 //remove left
26                 left++;
27 
28                 while (left < right)
29                 {
30                     if (s.charAt(left) != s.charAt(right)) return false;
31                     left++; right--;
32                 }
33             }
34         }
35         return true;
36     }
37 }

参考@yashar 的代码。

 

解法二：

 1 class Solution {
 2     public boolean validPalindrome(String s) 
 3     {
 4         return validPalindrome(s, 0, s.length() - 1, false);
 5     }
 6     
 7     private boolean validPalindrome(String s, int left, int right, Boolean mismatch)
 8     {
 9         if (right < left) return true;
10 
11         if (s.charAt(left) != s.charAt(right))
12         {
13             if (mismatch == true)
14                 return false;
15 
16             return validPalindrome(s, left, right - 1, true) || validPalindrome(s, left + 1, right, true);
17         }
18         else
19         {
20             return validPalindrome(s, left + 1, right - 1, mismatch);
21         }
22     }
23 }

递归，参考@yashar 的代码。

 

解法三：

 1 class Solution {
 2 public:
 3     bool validPalindrome(string s) {
 4         return valid(s, 0, s.length() - 1, 1);
 5     }
 6 
 7 private:
 8     bool valid(string& s, int i, int j, int d) { // d: num of chars you can delete at most
 9         if (i >= j) return true;
10         if (s[i] == s[j])
11             return valid(s, i + 1, j - 1, d);
12         else
13             return d > 0 && (valid(s, i + 1, j, d - 1) || valid(s, i, j - 1, d - 1));
14     }
15 };

递归，参考@alexander 的代码。