Note

Some notes need to remember:

1. Add values to some key in dict.(set default if this key not in dict)

   i. Method 1: Using dict.setdefault(key, default = None)

        >>> d = {}
        >>> for k, v in s:
        ...     d.setdefault(k, []).append(v)
        ...
        >>> d.items()
        [('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])]
   

   ii. Method 2: Using collections, which will be much faster than the above one:

        >>> s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
        >>> d = defaultdict(list)
        >>> for k, v in s:
        ...     d[k].append(v)
        ...
        >>> d.items()
        [('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])]
   

   ---

2. x >> 1: divide by 2
   x << 1: multiply by 2.

3. Reverse a Linked List:

    def getReverse(self, head):
        new_head = None
        while head:
            dummy = ListNode(head.val)
            dummy.next = new_head
            new_head = dummy
            head = head.next
        return new_head
   

4. n & (n-1): drops the lowest set bit.

    For example: n = 00101100
    1) n = 00101100, then n-1 = 00101011, so n & (n-1) = 00101000. Count = 1
    2) n = 00101000, then n-1 = 00100111, so n & (n-1) = 00100000. Count = 2
    3) n = 00100000, then n-1 = 00011111, so n & (n-1) = 00000000. Count = 3
   

   This way, we can count 1s in a number