标签：style blog http io color ar os for sp
Valid Palindrome
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.
左右分别往中间扫描，如果不同则返回false，如果两端碰面了则返回true
代码里写了很多注释，我认为这类题难点在于循环的判断条件以及跳出条件
因此对于每一个循环都需要明确它的进入条件以及跳出条件。
class Solution { public: bool isPalindrome(string s) { if(s.empty()) return true; else { int left = 0; int right = s.size()-1; //initilize c1 == c2, in case no move for left or right char c1 = ‘\0‘; char c2 = ‘\0‘; while(left < right) {//when left >= right, means true when left < right, jump out while(left < right) { if(s[left] >= ‘a‘ && s[left] <= ‘z‘) {//lowercase c1 = s[left]; break; } else if(s[left] >= ‘A‘ && s[left] <= ‘Z‘) {//capital c1 = tolower(s[left]); break; } else if(s[left] >= ‘0‘ && s[left] <= ‘9‘) {//number c1 = tolower(s[left]); break; } else // not a letter left ++; } //when come here, either left==right(1) or find next c1(2) //case (1): c1 and c2 remain unchanged and break out of while and return true //case (2): c1 get a new char if(left == right) //case (1) return true; //case (2) while(left <= right) {//careful! c2 must update the value as c1, even if the same position if(s[right] >= ‘a‘ && s[right] <= ‘z‘) {//lowercase c2 = s[right]; break; } else if(s[right] >= ‘A‘ && s[right] <= ‘Z‘) {//capital c2 = tolower(s[right]); break; } else if(s[right] >= ‘0‘ && s[right] <= ‘9‘) {//number c2 = tolower(s[right]); break; } else // not a letter right --; } //If come here, that means c1 and c2 are both updated if(c1 != c2) return false; else { left ++; right --; } } return true; } } };
标签：style blog http io color ar os for sp
原文地址：http://www.cnblogs.com/ganganloveu/p/4077246.html
