Oh My Leetcode
Leetcode 题解
**第 0000 题：**Power of Two
Given an integer, write a function to determine if it is a power of two.
**第 0001 题：**Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.
**第 0002 题：**First Missing Positive
Given an unsorted integer array, find the first missing positive integer.
For example, Given [1,2,0] return 3, and [3,4,-1,1] return 2.
Your algorithm should run in O(n) time and uses constant space.
**第 0003 题：**String to Integer (atoi)
Implement atoi to convert a string to an integer.
**第 0004 题：**Reverse Bits
Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).
**第 0005 题：**Palindrome Linked List
Given a singly linked list, determine if it is a palindrome.
Follow up: Could you do it in O(n) time and O(1) space?
**第 0006 题：**Sort Colors
Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Note: You are not suppose to use the library's sort function for this problem.
**第 0007 题：**LRU Cache
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
**第 0008 题：**Merge Two Sorted Lists
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.