Pointer

Pointer

It includes python, but not just python, but also C, Java, any programming language

Introduction

Pointers are an essential part of a programming language, and if you don’t use them correctly, you’re basically losing all the functionality and flexibility that programming languages allow.Pointer is not only a language foundation or characteristic, but also an idea. if you want learning well.

The pointer can be understood as a bridge of communication，So，It’s really important！

what’s Pointer?

A pointer is a variable which contains the address in memory of another variable. We can have a pointer to any variable type.

The unary or monadic operator & gives the ``address of a variable’’.

The indirection or dereference operator * gives the ``contents of an object pointed to by a pointer’’.

When and where to use it?

• When you want find anyone in tuple\List\Array\number from numbers\ string from str…
• Where you want to swap places
• When you want to sort
• In computer memory
• Method or object
• 。。。

Oh，It has so many uses

How to use it？

The use of the foundation can be found in other articles. I will use the pointer or pointer thought to accomplish something

Practical topics

Answer source code,Suggest doing it first

1. LeetCode-Two-sum
2. LeetCode-Three-sum
3. LeetCode-Four-sum
4. LeetCode-move-zeroes
5. LeetCode-reverse-linked-list
6. LeetCode-container-with-most-water
7. LeetCode-reverse-linked-list
8. LeetCode-swap-nodes-in-pairs
9. LeetCode-linked-list-cycle
10. LeetCode-linked-list-cycle-ii
11. … …

TwoSum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly\ one solution, and you may not use the same element twice. You can return the answer in any order.

# Method 1
'''
Tow Pointer,Nested loop.
'''
# Pseudo code
for loop -> i(nums.index)
for loop -> j(nums.index)
if nums[i] + num[j] == target and nums[i] != nums[j]:
# two pointer: i, j
return i, j
# Method 2
'''
Because of this relationship: nums[i] + nums[j] == target
therefore:  nums[i] = target - nums[j]
'''
# Pseudo code
for loop -> i (nums.index)
a = target - nums[i]
if a in nums and nums.index(a) != i:
return i, nums.index(a)
# Method 3 - other ideas
'''
Pursuing speed
You might think about the relationship between index and number
if you know dcit from key -> values find is fast, you maybe use it
'''
new HashMap
index -> Value - > index

Move Zore

Given an array nums, write a function to move all 0‘s to the end of it while maintaining the relative order of the non-zero elements. Note:

1. You must do this in-place without making a copy of the array.
2. Minimize the total number of operations.

# two pointers slow and fast: -> i, j
i -> start
j -> for loop
when j != 0, Move it to the front that from nums

Linked List Cycle

Given head, the head of a linked list, determine if the linked list has a cycle in it. There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to. Note that pos is not passed as a parameter. Return true if there is a cycle in the linked list. Otherwise, return false. Follow up: Can you solve it using O(1) (i.e. constant) memory? Constraints:

• The number of the nodes in the list is in the range [0, 104].
• -105 <= Node.val <= 105
• pos is -1 or a valid index in the linked-list.

# two pointer: slow and fast
when  slow == fast
return

Pointers are basic, but flexible. I believe you can also learn its flexible characteristics, as well as the idea of pointer. All changes do not depart from their ancestorsBelieve me, you can, too.