1、問題描述
在的時間複雜度和的空間複雜度內對鏈表進行排序。
例如:
輸入: 3->2->1->5->4
輸出:1->2->3->4->5
2、解題思路
- 思路1:最簡單的方法是利用選擇排序的思想,每次從鏈表中選擇第k大的數,插入表頭(頭節點的後面),k=1,2,…,n。這種方法的時間複雜度爲,空間複雜度爲,明顯不符合要求。
- 思路2:要求時間複雜度爲,類比數組的歸併排序,自然也會想到對鏈表採用歸併排序的方法,即將鏈表遞歸的劃分成兩部分,分別對每一部分進行排序,最後再將已經已排行序的兩部分合併成一部分。歸併算法主要涉及到cut和merge,cut是將鏈表不斷地切成兩部分,如果採用遞歸,則時間複雜度爲,空間複雜度也爲(因爲進行次調用),並不滿足題目要求,所以不能採用遞歸的方法;merge是將兩個排好序的鏈表連接成新的排序列表。下面介紹一種自底向上的歸併排序算法,它的基本思想如下:
- 這個方法主要用到鏈表中的兩個操作:
- (1)merge(list1,list2):二路歸併,把兩個有序的鏈表連接成一個新的有序鏈表;
- (2)cut(list,n):將鏈表list切掉前n個節點,並返回後半部分的鏈表頭;
- (3)dummyhead:創建一個新的節點作爲鏈表第一個節點的前置節點,這樣,就可以不用考慮鏈表頭節點是否爲空的情況了,這在刪除一個節點時十分重要,返回時只需要返回dummyhead->next,而不用存儲頭節點。
- 整個算法的思想如下:
輸入:鏈表
輸出:排好序的新鏈表
dummyhead->next = head;
for(j=1;j<list.length; j=j*2)
{
current = dummyhead->next;
tail = dummyhead
while (current !=null ){
left = current;
right = cut(current,step);
current = cut(right,step);
tail = merge(left,right)
while(tail->next != null) tail = tail->next;
}
}
以4->3->2->1->6->5爲例解釋該算法的求解過程:
3、代碼實現
- cut函數實現:
LinkNode* Cut(LinkNode* list, int node_nums){
LinkNode* dummyhead = new LinkNode;
dummyhead->next = list;
LinkNode* pointer = dummyhead;
while (pointer->next && node_nums){
pointer = pointer->next;
node_nums -= 1;
}
LinkNode* lefthead = pointer->next;
pointer->next = nullptr;
return lefthead;
}
- merge函數實現:
LinkNode* Merge(LinkNode* list1, LinkNode* list2){
LinkNode* p1=list1;
LinkNode* p2=list2;
LinkNode* dummyhead = new LinkNode;
LinkNode* iter=dummyhead;
while (p1 != nullptr && p2 != nullptr){
if (p1->value < p2->value){
iter->next = p1;
p1 = p1->next;
}
else{
iter->next = p2;
p2 = p2->next;
}
iter = iter->next;
}
while (p1 != nullptr){
iter->next = p1;
p1 = p1->next;
iter = iter->next;
}
while (p2 != nullptr){
iter->next = p2;
p2 = p2->next;
iter = iter->next;
}
return dummyhead->next;
}
SortedList函數實現:
LinkNode* SortedList(LinkNode* headpointer){
if (!headpointer || !headpointer->next){
return nullptr;
}
LinkNode* dummyhead = new LinkNode;
LinkNode* iter = headpointer->next;
int len = 0;
while (iter != nullptr){
len += 1;
iter = iter->next;
}
dummyhead->next = headpointer->next;
LinkNode* current = nullptr;
LinkNode* tail = nullptr;
for (int step = 1; step < len; step *= 2){
current = dummyhead->next;
tail = dummyhead;
while (current != nullptr){
LinkNode* left = current;
LinkNode* right = Cut(current, step);
current = Cut(right, step);
tail->next = Merge(left, right);
while (tail->next){
tail = tail->next;
}
}
}
return dummyhead->next;
}
- LinkList.h
struct LinkNode{
int value;
LinkNode* next=nullptr;
};
class LinkList{
public:
LinkNode* headpointer;
LinkList(int* value_array,int len);
void PrintLinkList();
};
- LinkList.cpp
#include"LinkList.h"
#include<iostream>
using namespace std;
LinkList::LinkList(int* array_value,int len){
//LinkNode headnode;
//LinkNode* pointer = &headnode;
LinkNode* headnode = new LinkNode();
LinkNode* pointer = headnode;
//int len = sizeof(array_value) / sizeof(int);
for (int i = 0; i < len; i++){
LinkNode* newnode = new LinkNode();
newnode->value = array_value[i];
pointer->next = newnode;
pointer = pointer->next;
}
this->headpointer = headnode;
}
void LinkList::PrintLinkList(){
LinkNode* pointer = this->headpointer->next;
while (pointer->next != nullptr){
cout << pointer->value << "->";
pointer = pointer->next;
}
cout <<pointer->value<<endl;
}
- 主函數
#include<iostream>
#include"LinkList.h"
using namespace std;
//#define test
int main(){
int values[] = {4,3,2,1,6,5};
LinkList* list = new LinkList(values,sizeof(values)/sizeof(values[0]));
#ifdef test
LinkNode* pointer = list->headpointer->next;
while (pointer != nullptr){
cout << pointer->value << "-->";
pointer = pointer->next;
}
cout << endl;
#else
cout << "before sorting:" << endl;
list->PrintLinkList();
cout << "after sorting:" << endl;
LinkNode* result = SortedList(list->headpointer);
LinkNode* pointer = result;
while (pointer && pointer->next != nullptr){
cout << pointer->value << "->";
pointer = pointer->next;
}
cout << pointer->value << endl;
#endif
return 0;
}