文章列表
3 primary features of OOP
Encapsulation
Inheritance
Polymorphism
class
A user-defined datatype which groups together related pieces of information
Instance
An instance is an occurrence of a class
class student{
public:
char *name;
int studentID;
};
int main(){
student s1;
student s2;
}
Constructor
Method that is calle whien ...
&x evaluates to the adress of x in memory.
*(&x) evaluates to the someting as x.
Pointer advantages:
More flexible pass-by-reference
Manipulate comlex data structure efficiently
Use polymorphism
Pointers are variables stroring memory address.
Declaring pointers:
int *ptr = &a ...
Array
fixed number of elements of the same type stored sequentially in memory
initialization, or unexpected results
arrays are passed by reference
multidimensional arrays are merely an abstraction for programmers, as all the elements in the array are sequential in memory
String
simp ...
Why define your own functions?
Readability
Maintainability
Code reuse
#include <iostream>
using namespace std;
int raiseToPower(int base, int exponent) {
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return result;
}
int main( ...
Control structure
conditionals
if
switch
loops
while
do...while
for
nested control structures
这些都跟java是一致的，谁叫java是从C++改过来的呢？
1.1 Why use a language like C++
Conciseness
Maintainability
Portability
1.2 The compilation process
sorce file>>>Prcessed code>>>Object file
1.3 General notes on C++
case sensitive
2.1 hello world
/*
* File: main.cpp
* Author: Gong ...
class Node<E>{
E item;
Node prev, next;
Node(E newItem){
item = newItem;
prev = next =null;
}
}
class CircularDoublyLinkedList{
private Node head;
private Node tail;
private int size;
CircularDoublyLinkedList(){
...
Let us consider the below traversals:
Inorder sequence: D B E A F CPreorder sequence: A B D E C F
In a Preorder sequence, leftmost element is the root of the tree. So we know ‘A’ is root for given sequences. By searching ‘A’ in Inorder sequence, we can find out all elements on left side of ...
二叉树和图：确定能否从一个点遍历到另一个节点
- 博客分类：
- 数据结构
Use recursion and traversal to solve.
boolean BSTbranch(TreeNode start, TreeNode end){
if(start == end) return true;
if(start != null){
if(!BSTbranch(start.left, end){
if(!BSTbranch(start.right, end)
return false;
else
return true;
...