【leetcode c++】125 Valid Palindrome
2015-10-09 17:38
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Valid Palindrome
Given a string, determine if it is apalindrome, considering only alphanumeric characters and ignoring cases.
For example,
"A man, a plan, a canal: Panama" isa palindrome.
"race a car" is not a palindrome.
Note:
Have you consider that the string might beempty? This is a good question to ask during an interview.
For the purpose of this problem, we defineempty string as valid palindrome.
字有点多,题目要求我们判断一个字符串中的字母&数字是不是按回文的顺序出现的,也就是,把这个字符串中的字母&数字提取出来看它是不是回文的。
空串呢?这是个好问题,我们约定空串是回文的。
当然,我们肯定不去提取字母&数字,我们用迭代器去从头和尾向中间遍历。遇到字母&数字,就停下判断,非字母&数字就向中靠。扫描完字符串为止。
class Solution {
public:
bool isPalindrome(string s) {
string::iterator iter1 = s.begin();
string::iterator iter2 = s.end();
iter2--;
while(iter1 < iter2)
{
if(!isalnum(*iter1))//iter1所指不是字母&数字,向中
{
iter1++;
continue;
}
else if(!isalnum(*iter2))//iter2所指不是字母&数字,向中
{
iter2--;
continue;
}
//到这里,iter1 iter2所指都是字母&数字
if((*iter1 == *iter2) || 32 == abs(*iter1 - * iter2))//同字符,继续向中
{
iter1++;
iter2--;
continue;
}
else return false;//字符不同,肯定不是回文。
}
return true;//扫描完毕,没发现非法,是回文
}
};
Given a string, determine if it is apalindrome, considering only alphanumeric characters and ignoring cases.
For example,
"A man, a plan, a canal: Panama" isa palindrome.
"race a car" is not a palindrome.
Note:
Have you consider that the string might beempty? This is a good question to ask during an interview.
For the purpose of this problem, we defineempty string as valid palindrome.
字有点多,题目要求我们判断一个字符串中的字母&数字是不是按回文的顺序出现的,也就是,把这个字符串中的字母&数字提取出来看它是不是回文的。
空串呢?这是个好问题,我们约定空串是回文的。
当然,我们肯定不去提取字母&数字,我们用迭代器去从头和尾向中间遍历。遇到字母&数字,就停下判断,非字母&数字就向中靠。扫描完字符串为止。
class Solution {
public:
bool isPalindrome(string s) {
string::iterator iter1 = s.begin();
string::iterator iter2 = s.end();
iter2--;
while(iter1 < iter2)
{
if(!isalnum(*iter1))//iter1所指不是字母&数字,向中
{
iter1++;
continue;
}
else if(!isalnum(*iter2))//iter2所指不是字母&数字,向中
{
iter2--;
continue;
}
//到这里,iter1 iter2所指都是字母&数字
if((*iter1 == *iter2) || 32 == abs(*iter1 - * iter2))//同字符,继续向中
{
iter1++;
iter2--;
continue;
}
else return false;//字符不同,肯定不是回文。
}
return true;//扫描完毕,没发现非法,是回文
}
};
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