Leetcode 125 验证回文串 双指针

　　双指针解法1：

    public final boolean isPalindrome(String s) {
        int length = s.length();
        int leftPoint = 0;
        int rightPoint = length - 1;
        s = s.replaceAll("[\pP‘’“”``]", " ");
        while (leftPoint < rightPoint) {
            char leftChar = s.charAt(leftPoint);
            while (leftPoint < rightPoint && leftChar == ' ') {
                leftPoint++;
                leftChar = s.charAt(leftPoint);
            }
            char rightChar = s.charAt(rightPoint);
            while (leftPoint < rightPoint && rightChar == ' ') {
                rightPoint--;
                rightChar = s.charAt(rightPoint);
            }
            if (leftPoint < rightPoint) {
                leftChar = Character.toLowerCase(leftChar);
                rightChar = Character.toLowerCase(rightChar);
                if (leftChar != rightChar) {
                    return false;
                }
                leftPoint++;
                rightPoint--;
            }
        }
        return true;
    }

　　双指针解法2：

    public boolean isPalindrome0(String s) {
        int n = s.length();
        int left = 0, right = n - 1;
        while (left < right) {
            while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
                ++left;
            }
            while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
                --right;
            }
            if (left < right) {
                if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
                    return false;
                }
                ++left;
                --right;
            }
        }
        return true;
    }

　　两个解法都是双指针，区别在于方法  1 使用正则表达式一次性过滤了所有符号字符；而方法二通过 Character.isLetterOrDigit 方法逐个判断每个字符是不是标点。

　　方法二为懒处理，遇到了才处理，对于可能在任何位置直接返回 false 的函数来说，综合开销小于方法 1 。