leetcode 125. Valid Palindrome | 回文string

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;}

• 别人的代码主要是有isalnum()/toupper()精简,基本逻辑一致.
• 另外i/j的两个while也很漂亮

参考

• isalnum 函数
• toupper 函数