Valid Palindrome
- tags: [palindrome]
Question
- leetcode: Valid Palindrome | LeetCode OJ
- lintcode: (415) Valid Palindrome
Given a string, determine if it is a palindrome,
considering only alphanumeric characters and ignoring cases.
Example
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.
Note
Have you consider that the string might be empty?
This is a good question to ask during an interview.
For the purpose of this problem,
we define empty string as valid palindrome.
Challenge
O(n) time without extra memory.
題解
字符串的回文判斷問題,由於字符串可隨機訪問,故逐個比較首尾字符是否相等最為便利,即常見的『兩根指針』技法。此題忽略大小寫,並只考慮字母和數字字符。鏈表的回文判斷總結見 Check if a singly linked list is palindrome.
Python
class Solution:
# @param {string} s A string
# @return {boolean} Whether the string is a valid palindrome
def isPalindrome(self, s):
if not s:
return True
l, r = 0, len(s) - 1
while l < r:
# find left alphanumeric character
if not s[l].isalnum():
l += 1
continue
# find right alphanumeric character
if not s[r].isalnum():
r -= 1
continue
# case insensitive compare
if s[l].lower() == s[r].lower():
l += 1
r -= 1
else:
return False
#
return True
C++
class Solution {
public:
/**
* @param s A string
* @return Whether the string is a valid palindrome
*/
bool isPalindrome(string& s) {
if (s.empty()) return true;
int l = 0, r = s.size() - 1;
while (l < r) {
// find left alphanumeric character
if (!isalnum(s[l])) {
++l;
continue;
}
// find right alphanumeric character
if (!isalnum(s[r])) {
--r;
continue;
}
// case insensitive compare
if (tolower(s[l]) == tolower(s[r])) {
++l;
--r;
} else {
return false;
}
}
return true;
}
};
Java
public class Solution {
/**
* @param s A string
* @return Whether the string is a valid palindrome
*/
public boolean isPalindrome(String s) {
if (s == null || s.isEmpty()) return true;
int l = 0, r = s.length() - 1;
while (l < r) {
// find left alphanumeric character
if (!Character.isLetterOrDigit(s.charAt(l))) {
l++;
continue;
}
// find right alphanumeric character
if (!Character.isLetterOrDigit(s.charAt(r))) {
r--;
continue;
}
// case insensitive compare
if (Character.toLowerCase(s.charAt(l)) == Character.toLowerCase(s.charAt(r))) {
l++;
r--;
} else {
return false;
}
}
return true;
}
}
源碼分析
兩步走:
- 找到最左邊和最右邊的第一個合法字元(字母或者字元)
- 一致轉換為小寫進行比較
字元的判斷盡量使用語言提供的 API
複雜度分析
兩根指標遍歷一次,時間複雜度 , 空間複雜度 .