Palindrome Number 回文数字
Determine whether an integer is a palindrome. Do this without extra space.
前期处理，首先负数/10的倍数，直接return false；
然后【1】直观做法是，知道N位数，根据value/10^(N-i)和value%10^i来判断，然后value=value%10^(N-i)/10^i，其中i = 1,..., N/2 【2】Brilliant做法是，通过value的最后一位，作为新数的当前位，循环操作，直到value/=10更新后的value <= 新数，复杂度更低，return value == 新数
Valid Palindrome 回文字符串
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
两端下标，往中间靠近，判断是否相同字母（其中大小写相差d = 'A' - 'a'，根据'A~Z'，'a~z'范围来判断+-d）
Palindrome Linked List 回文链表
Given a singly linked list, determine if it is a palindrome. O(1) space
//Definition for singly-linked list.
struct ListNode {
int val;
struct ListNode *next;
};
要求O(1)空间复杂度，通过分割、翻转、双指针遍历完成
Palindrome Permutation I
Given a string, determine if a permutation of the string could form a palindrome.
For example,"code" -> False, "aab" -> True, "carerac" -> True.
用一个数组（或者hash）计数，如果奇数数目的字母个数大于1个，则return false；否则return true。时间复杂度为O(N)遍历一次+O(1)遍历一次 = O(N)。
Palindrome Permutation II
Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.
For example:
Given s = "aabb", return ["abba", "baab"].
Given s = "abc", return [].
跟1一样的思想，由于要全部输出，就DFS回溯