给定一个整数，编写一个函数来判断它是否是 2 的幂次方。
示例 1:
输入: 1
输出: true
解释: 20 = 1
class Solution {
public boolean isPowerOfTwo(int n) {
if (n == 0)
return false;
long x = (long)n;
return (x & -x) == x;
}
}
请判断一个链表是否为回文链表。
示例 1:
输入: 1->2
输出: false
class Solution {
public boolean isPalindrome(ListNode head) {
if (head == null) {
return true;
}
ListNode left = endOfFirstHalf(head);
ListNode right = reverseNode(left.next);
ListNode p1 = head;
ListNode p2 = right;
boolean ans = true;
while (ans && p2 != null) {
if (p1.val != p2.val)
ans = false;
p1 = p1.next;
p2 = p2.next;
}
left.next = reverseNode(right);
return ans;
}
private ListNode reverseNode(ListNode head) {
ListNode pre = null;
ListNode cur = head;
while (cur != null) {
ListNode temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
}
return pre;
}
private ListNode endOfFirstHalf(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while (fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
}
给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。
给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。示例 1：
输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]
class Solution {
public List<String> letterCombinations(String digits) {
List<String> combinations = new ArrayList<>();
if (digits.length() == 0)
return combinations;
Map<Character, String> phoneMap = new HashMap<>() {
{
put('2', "abc");
put('3', "def");
put('4', "ghi");
put('5', "jkl");
put('6', "mno");
put('7', "pqrs");
put('8', "tuv");
put('9', "wxyz");
}
};
backtrack(combinations, phoneMap, digits, 0, new StringBuffer());
return combinations;
}
public void backtrack(List<String> combinations, Map<Character, String> phoneMap, String digits, int index,
StringBuffer combination) {
if (index == digits.length()) {
combinations.add(combination.toString());
} else {
char digit = digits.charAt(index);
String letters = phoneMap.get(digit);
int lettersCount = letters.length();
for (int i = 0; i < lettersCount; i++) {
combination.append(letters.charAt(i));
backtrack(combinations, phoneMap, digits, index + 1, combination);
combination.deleteCharAt(index);
}
}
}
}