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c++ - 链表排序不影响 main (C++) 中的列表项?

转载 作者:太空宇宙 更新时间:2023-11-04 13:17:23 25 4
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我正在尝试根据点到原点的距离对点的链接列表进行排序。点越近,应该是链表中的第一个/下一个。示例:如果用户输入 (2,2) (4,4) (1,1) (3,3),则应使用指向 (1,1) (2,2) (3, 3)(4,4)。它的工作原理除了 main 中的点不受排序的影响并且点不互换,所以如果 (1,1) 像示例中的那样是第三个,那么与(4,4).

#include <iostream>
#include <math.h>
using namespace std;

class List {
public:
int x;
int y;
List *next;
};
void init(List *root) {
int x, y;
List *traverse;
traverse = root;
while(traverse != 0) {
cout<<"Enter coordinate x: ";
cin>>x;
traverse->x = x;
cout<<"Enter coordinate y: ";
cin>>y;
traverse->y = y;
traverse = traverse->next;
}
}
void display(List *root) {
List *traverse;
traverse = root;
while(traverse->next != 0) {
cout<<"("<<traverse->x<<","<<traverse->y<<") ";
traverse = traverse->next;
}
cout<<"("<<traverse->x<<","<<traverse->y<<")"<<endl;
}
void sort(List *root, int n) {
List *traverse;
List *stable;
traverse = root;
stable = root;
double d1, d2;
for(int i = 0; i < n; i++) {
d1 = sqrt(pow(traverse->x, 2) + pow(traverse->y, 2));
d2 = sqrt(pow(stable->x, 2) + pow(stable->y, 2));
if(d1 < d2) {
root = traverse;
}
traverse = traverse->next;
}
}
void collinear(List *root) {
int x[3] = {0}, y[3] = {0};
int value;
List *traverse;
traverse = root;
for(int i = 0; i < 3; i++) {
if(traverse != 0) {
x[i] = traverse->x;
y[i] = traverse->y;
traverse = traverse->next;
}
}
if(x[2] != 0) {
value = x[0] * (y[1] - y[2]) + x[1] * (y[2] - y[0]) + x[2] * (y[0] - y[1]);
if(value == 0) {
traverse = root;
for(int i = 0; i < 3; i++) {
cout<<"("<<traverse->x<<","<<traverse->y<<") ";
traverse = traverse->next;
}
cout<<"collinear!"<<endl;
}
else {
traverse = root;
for(int i = 0; i < 3; i++) {
cout<<"("<<traverse->x<<","<<traverse->y<<") ";
traverse = traverse->next;
}
cout<<"non-collinear!"<<endl;
}
}
else {
cout<<"Not a group of 3 points cannot calculate collinearity!"<<endl;
}
}
int main() {
List *root;
root = new List;
List *traverse;
traverse = root;
List *node1;
node1 = new List;
List *node2;
node2 = new List;
List *node3;
node3 = new List;
root->next = node1;
node1->next = node2;
node2->next = node3;
node3->next = 0;
init(traverse);
display(traverse);
/*for(int i = 0; i < 2; i++) {
collinear(traverse);
traverse = traverse->next;
}
traverse = root;*/
traverse = root;
int n = 4;
for(int i = 0; i < 4; i++) {
List *traverse2;
traverse2 = traverse;
sort(traverse2, n);
traverse = traverse->next;
n--;
}
traverse = root;
display(traverse);
return 0;
}

最佳答案

如果您希望能够更改根目录,您需要将其作为 ** 传递。然而,即使那样我也不认为你的排序函数会起作用,它只会返回一个截断的列表,在最小元素之前缺少所有内容。

您使用自定义链表和排序算法是否有任何原因?标准库负责处理诸如容器和排序之类的事情:

#include <vector>
#include <algorithm>
#include <iostream>

class Point
{
public:
Point(int x_, int y_) :
x(x_),
y(y_)
{
}

// no need to square root to compare distance from origin
int distanceSquared() const
{
return (x*x)+(y*y);
}

// define an operator< so we can sort containers holding this class
bool operator< ( const Point& rhs ) const
{
return distanceSquared() < rhs.distanceSquared();
}

// declare a friend function for outputing the values in a user friendly way
friend std::ostream& operator<<(std::ostream& os, const Point& p);

int x;
int y;
};

// and define the friend function outside the class
std::ostream& operator<<(std::ostream& os, const Point& p)
{
os << "(" << p.x << ", " << p.y << ")";
return os;
}

int main()
{
// std::vector to store user entered points
std::vector<Point> points;
// temp storage for user input
int x, y;
// entering anything non-numberic will exit this loop
while( true )
{
std::cout << "Enter X coordinate : ";
std::cin >> x;
if( std::cin.fail() )
break;
std::cout << "Enter Y coordinate : ";
std::cin >> y;
if( std::cin.fail() )
break;
// we've read in a valid x,y so add a new point to vector
points.push_back( Point(x, y) );
}

// sort will use the operator< function in Point class
std::sort(points.begin(), points.end());

// use the stream operator to debug out our point
for( auto it = points.begin(); it != points.end(); ++it )
{
std::cout << *it << std::endl;
}
}

关于c++ - 链表排序不影响 main (C++) 中的列表项?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36702312/

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