Valid Palindrome
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
Note:
For the purpose of this problem, we define empty string as valid palindrome.
Example 1:
Input:
"A man, a plan, a canal: Panama"
Output:
true
Example 2:
Input: "race a car"
Output: false
//问题：回文（判断一个字符串是否是回文的，这里仅考虑字母数字字符，且忽略大小写）
//方法：双指针法，分别从开头和结尾扫描
using
namespace
std
;
#include <locale>
//本地化库的一部分，包含字符分类、转换等函数
class
Solution
{
public
:
bool
isPalindrome
(
string s
)
{
for
(
int
i
=
0
,
j
=
s
.
size
()-
1
;
i
<
j
;
i
++,
j
--)
//双指针，分别从开头和结尾开始扫描
{
while
(
isalnum
(
s
[
i
])
==
false
&&
i
<
j
)
i
++;
//如果不是字母数字字符(alphanumeric),增加左指针
while
(
isalnum
(
s
[
j
])
==
false
&&
i
<
j
)
j
--;
//如果不是字母数字字符(alphanumeric),增加右指针
if
(
toupper
(
s
[
i
])
!=
toupper
(
s
[
j
]))
return
false
;
//如果不匹配就退出
}
return
true
;
}
};
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.
//问题：回文2（判断一个字符串是否是回文，可以最多删除一个字符，而且字符串中只有小写英文字母，最大长度为50000）
//方法：双指针法，借用回文1的解法
#include <iostream>
class
Solution
{
public
:
bool
validPalindrome
(
string s
)
{
for
(
int
i
=
0
,
j
=
s
.
size
()-
1
;
i
<
j
;
i
++,
j
--)
//双指针，分别从开头和结尾开始扫描
{
if
(
s
[
i
]
!=
s
[
j
]) //扫描到不匹配字符时，删除其中一个，然后继续扫描
{
int
i1
=
i
,
j1
=
j
-
1
;
//“删除”右边元素
int
i2
=
i
+
1
,
j2
=
j
;
//“删除”左边元素
while
(
i1
<
j1
&&
s
[
i1
]
==
s
[
j1
])
//继续扫描剩余元素
{
i1
++;
j1
--;
}
while
(
i2
<
j2
&&
s
[
i2
]
==
s
[
j2
])
{
i2
++;
j2
--;
}
return
i1
>=
j1
||
i2
>=
j2
;
//i1>=j1代表已经扫描完毕，字母均匹配
}
}
return
true
;
}
};