leetcode 125. Valid Palindrome | 回文string-优快云博客

本文链接：https://blog.youkuaiyun.com/Einsteinzhao/article/details/75670198

Description

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.

My solution

// 我的解法太长了,显然是可以被精简的,里面太多if规则
class Solution {
public:
    bool isPalindrome(string s) {
        int i = 0, j = s.size() - 1;
        int iscapi, iscapj;
        while (i < j) {
            iscapi = 0;
            iscapj = 0;
            for (; i < j;) {
                if ('a' <= s[i] && s[i] <= 'z') {
                    iscapi = 1;
                    break;
                }
                if ('A' <= s[i] && s[i] <= 'Z') {
                    iscapi = 2;
                    break;
                }
                if ('0' <= s[i] && s[i] <= '9') break;
                i++;
            }
            for (; i < j;) {
                if ('a' <= s[j] && s[j] <= 'z') {
                    iscapj = 1;
                    break;
                }
                if ('A' <= s[j] && s[j] <= 'Z') {
                    iscapj = 2;
                    break;
                }
                if ('0' <= s[j] && s[j] <= '9') break;
                j--;
            }
            if(i==j) break;
            if (iscapi == iscapj) {
                if (s[i] != s[j]) return false;
            } else {
                if (iscapi == 2)
                    if (s[i] - s[j] != -32) return false; else;
                else if (s[i] - s[j] != 32) return false;
            }
            i++;
            j--;
        }
        return true;
    }
};

当心上面有悬挂else陷阱.之前发生了错误.

Discuss

bool isPalindrome(string s) {
    for (int i = 0, j = s.size() - 1; i < j; i++, j--) { // Move 2 pointers from each end until they collide
        while (isalnum(s[i]) == false && i < j) i++; // Increment left pointer if not alphanumeric
        while (isalnum(s[j]) == false && i < j) j--; // Decrement right pointer if no alphanumeric
        if (toupper(s[i]) != toupper(s[j])) return false; // Exit and return error if not match
    }

    return true;
}