Bit Manipulation

Problem Names	Solution Writeups	Code
Single Number	Since the question gurantees that there is only one number that does not have its duplicates, the rest of the numbers all repeated twice. And we know that a XOR a = 0, so if we XOR all the numbers in the array, we will finally obtain that single number.
Number of 1 Bits	The input is fixed length of 32. Create a mask initially = 1, and then keep left shift by 1 to check each position contains 1 or 0.
Counting Bits	For each number generated, count how many 1 it holds. Record te result in an array.
Reverse Bits	Since input is a fixed size 32 bits integer, the way to reverse is that we check from rightmost everytime if that digit holds 1, ans += 1 << (31 - i), where i is the bit position.
Missing Number	One simple approach is that we first calculate cumulative sum of n, then for each number in the array, sum -= num. In the end sum will store the only missing number. Another approach of bit manipulation is that we first calculate the result after XOR all the numbers from 0 to n. Then in the given array, we again XOR result with num, and eventually we will get the only missing number.
Sum of Two Integers	Loop while b is not 0, Sum bit = a XOR b, carry bit = (unsigned int)(a&b)<<i, since when both digits contain 1, carry = 1. Then update a to sum_bit and b to carry.
Reverse Integer	Non-bit-manipulation approach is to push each digit to stack, and when popping from stack, times 10 to its corresponding power to the digit.