力扣https://leetcode-cn.com/problems/valid-palindrome/solution/yan-zheng-hui-wen-chuan-by-leetcode-solution/

此题比较简答，因为我做出来了，哈哈

思考总结

• 字符相等怎么判断相等，ASSIC码 

• 需要跳过不考核的字符
• 双指针，相向而行

package com.company.myarrays;

public class Solution_5 {

    /**
     * "0P"
     */
    public boolean isPalindrome(String s) {
        int left = 0, right = s.length() - 1;

        while (left < right) {
            while (!isCharAndNum(s.charAt(left)) && left < right) {
                left++;
            }
            char leftCh = s.charAt(left);

            while (!isCharAndNum(s.charAt(right)) && right > left) {
                right--;
            }
            char rightCh = s.charAt(right);


            if (isEqualChIgnoreCase(leftCh, rightCh)) {
                left++;
                right--;
                continue;
            } else {
                return false;
            }

        }
        return true;
    }

    private boolean isEqualChIgnoreCase(char leftCh, char rightCh) {
        //任意一个是数字
        if ((leftCh >= '0' && leftCh <= '9') || (rightCh >= '0' && rightCh <= '9')) {
            if (leftCh == rightCh) {
                return true;
            } else {
                return false;
            }
        }

        //两个都是字母
        if (leftCh - rightCh == 32 || leftCh - rightCh == -32 || leftCh == rightCh) {
            return true;
        }
        return false;
    }

    public boolean isCharAndNum(char ch) {
        if ((ch >= 'a' && ch <= 'z')
                || (ch >= 'A' && ch <= 'Z')
                || (ch >= '0' && ch <= '9')) {
            return true;
        } else {
            return false;
        }
    }

    public static void main(String[] args) {
        System.out.println('0'==48);
    }
}