🌗 125. 验证回文串

难度: 🌗

问题描述

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解法

class Solution {
    public boolean isPalindrome(String s) {
        // 思路：
        // 双指针
        int len = s.length();
        if(len <= 1) {
            return true;
        }
        // len >= 2
        int left = 0;
        int right = len - 1;
        int diff = Math.abs('A' - 'a');
        while(left < right) {
            while(left < right && !isDigit(s.charAt(left)) && !isValid(s.charAt(left))) {
                left ++;
            }
            while(left < right && !isDigit(s.charAt(right)) && !isValid(s.charAt(right))) {
                right --;
            }
            if(left == right) {
                return true;
            } else {
                // left < right
                char a = s.charAt(left);
                char b = s.charAt(right);
                if(isDigit(a) && a == b) {
                    left ++;
                    right --;
                } else if(isValid(a) && isValid(b)) {
                    if(a == b || Math.abs(a - b) == diff) {
                        left ++;
                        right --;
                    } else {
                        return false;
                    }
                } else {
                    return false;
                }
            }
        }
        return true;
    }

    private boolean isValid(char c) {
        if(c >= 'a' && c <= 'z') {
            return true;
        }
        if(c >= 'A' && c <= 'Z') {
            return true;
        }
        return false;
    }

    private boolean isDigit(char c) {
        if(c >= '0' && c <= '9') {
            return true;
        }
        return false;        
    }
}

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输出