知识点 1. 整数的二进制表示法 2. 十进制和二进制的转换 http://baike.baidu.com/view/1426817.htm 3. 负整数的表示(原码,补码,反码) http://www.cnblogs.com/zhangziqiu/archive/2011/03/30/ComputerCode.html 4. 位操作 Bit Operation 左移 Left Shift      << 右移 Right Shift    >> 与 And   & 或  …

Description Count how many 1 in binary representation of a 32-bit integer. Example Given 32, return 1 Given 5, return 2 Given 1023, return 9 Challenge If the integer is n bits with m 1 bits. Can you do it in O(m) time? 解题:很简单,但是要考虑范围的问题.代码如下: public…

题目: 二进制中有多少个1 49% 通过 计算在一个 32 位的整数的二进制表式中有多少个 1. 样例 给定 32 (100000),返回 1 给定 5 (101),返回 2 给定 1023 (111111111),返回 9 解题: Java程序: public class Solution { /** * @param num: an integer * @return: an integer, the number of ones in num */ public int countOnes…

Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) and an query list. For each query, give you an integer, return the number of element in the array that are smaller than the given integer. Have…

描述 设计一个算法,并编写代码来序列化和反序列化二叉树.将树写入一个文件被称为“序列化”,读取文件后重建同样的二叉树被称为“反序列化”. 如何反序列化或序列化二叉树是没有限制的,你只需要确保可以将二叉树序列化为一个字符串,并且可以将字符串反序列化为原来的树结构. 对二进制树进行反序列化或序列化的方式没有限制,LintCode将您的serialize输出作为deserialize的输入,它不会检查序列化的结果. 样例 给出一个测试数据样例, 二叉树{3,9,20,#,#,15,7},表示如下的树结…

Count how many 1 in binary representation of a 32-bit integer. Example Given 32, return 1 Given 5, return 2 Given 1023, return 9 public class Solution { /** * @param num: an integer * @return: an integer, the number of ones in num */ public int count…

Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Have you met this question in a real interview? Yes Example Given a binary tree as follow:…

The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. 11 is read off as "two 1s" or 21. 21 is read off as "one 2, then one 1" or 1211. Given…

1 /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root…

iven a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.…