Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For 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.
-----------------------------------------------------------
class Solution {
public:
bool isPalindrome(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
string str="";
for (int i=0;i<s.length();i++) {
if (isalnum(s[i])) {
str += tolower(s[i]);
}
}
int length = str.length();
for (int i=0;i<length/2;i++) {
if (str[i] != str[length - i-1]) {
return false;
}
}
return true;
}
};
Update on Jan-16-2015 constant space
class Solution {
public:
bool isPalindrome(string s) {
int left = 0;
int right = s.length();
while (left < right) {
if (isalnum(s[left]) && isalnum(s[right])) {
if (s[left] == s[right]
|| s[left] + 32 == s[right]
|| s[left] == s[right] + 32) {
left++;
right--;
}else {
return false;
}
}else if (!isalnum(s[left])) {
left++;
}else if (!isalnum(s[right])) {
right--;
}
}
return true;
}
};
Sum Root to Leaf Numbers
Given a binary tree containing digits from
0-9 only, each root-to-leaf path could represent a number.An example is the root-to-leaf path
1->2->3 which represents the number 123.Find the total sum of all root-to-leaf numbers.
For example,
1 / \ 2 3The root-to-leaf path
1->2 represents the number 12.The root-to-leaf path
1->3 represents the number 13.Return the sum = 12 + 13 =
25.-----------------------------------------------------------------
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void leef (TreeNode *root, int pathSum, int & sum) {
int sumL = pathSum, sumR = pathSum;
if (root->left != NULL) {
sumL = sumL*10 + root->left->val;
leef(root->left,sumL,sum);
}
if (root->right != NULL) {
sumR = sumR*10 + root->right->val;
leef(root->right,sumR,sum);
}
if (root->left == NULL && root->right == NULL) {
sum += pathSum;
}
}
int sumNumbers(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (root == NULL) return 0;
int sum=0;
leef(root,root->val,sum);
return sum;
}
};
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