标签:ble ide isa problem size cas empty eric val
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
思路:
设置两个指针,分别从前往后和从后往前遍历,如果遇到非数字或字母字符就跳过,这里可以用isalnum()函数进行判断,非常方便,然后判断2个为数字或字母的字符是否相等。注意大小写也认为是相同的,所以可以用toupper()函数将小写都变为大写再进行比较。
Solution:
bool isPalindrome(string s) {
for (int i = 0, j = s.size()-1; i < j; i++, j--) {
while (!isalnum(s[i]) && i < j) i++;
while (!isalnum(s[j]) && i < j) j--;
if (toupper(s[i]) != toupper(s[j])) return false;
}
return true;
}
性能:
Runtime: 8 ms??Memory Usage: 9.5 MB
标签:ble ide isa problem size cas empty eric val
原文地址:https://www.cnblogs.com/dysjtu1995/p/12267031.html