相關題解：
leetcode_效率題解_[python/C++]_147. Insertion Sort List（鏈表插入排序）
題目鏈接
【題目】
Sort a linked list in O(n log n) time using constant space complexity.
【分析】
O(nlogn)的複雜度我們很顯然想到歸併排序，快速排序
記得大一的時候寫鏈表排序的題都是先轉成數組或者vector然後再進行排序，貼出來以便大家熟悉一下數組的歸併排序
class Solution {
public:
ListNode* sortList(ListNode* head) {
if( head == NULL ) return head;
vector<int> vec;
while( head != NULL ){
vec.push_back(head->val);
head = head->next;
}
int size = vec.size();
merge_sort( vec , 0 , size - 1 );
ListNode * List = new ListNode(vec[0]);
ListNode * ans = List;
for( int i = 1 ; i < size ; i ++ ){
ListNode * temp = new ListNode(vec[i]);
ans->next = temp;
ans = ans->next;
}
return List;
}
void merge( vector<int>& a , int low , int mid , int high ){
vector<int> temp( high - low + 1 , 0 );
int i = low , j = mid+1 , m = mid , n = high , k = 0;
while( i <= m && j <= n ){
if( a[i] < a[j] ) temp[k++] = a[i++];
else temp[k++] = a[j++];
}
while( i <= m ) temp[k++] = a[i++];
while( j <= n ) temp[k++] = a[j++];
for( int g = 0 ; g < k ; g ++ ) a[low+g] = temp[g];
}
void merge_sort( vector<int>& a , int low , int high ){
if( low < high ){
int mid = ( low + high )/2;
merge_sort( a , low , mid );
merge_sort( a , mid+1 , high );
merge( a , low , mid , high );
}
}
};然而效率是這樣的
所以歸根到底我們還是得用指針來寫：
那重點在哪裏呢，在於如何找到一個鏈表的mid
這裏我用了下面一段代碼來找到分界點second_head
ListNode * slow = head;
ListNode * fast = head->next;
while(fast&&fast->next){
fast = fast->next->next;
slow = slow->next;
}
ListNode * second_head = slow->next;
slow->next = NULL;所以就簡單了：
class Solution {
public:
ListNode* sortList(ListNode* head) {
if( !head || !head->next ) return head;
ListNode * slow = head;
ListNode * fast = head->next;
while(fast&&fast->next){
fast = fast->next->next;
slow = slow->next;
}
ListNode * second_head = slow->next;
slow->next = NULL;
return merge( sortList(head) , sortList(second_head) );
}
ListNode * merge( ListNode * left , ListNode * right ){
ListNode * new_list = new ListNode(0);
ListNode * temp = new_list;
while( left && right ){
if( left->val < right->val ){
temp=temp->next = left;
left = left->next;
}
else{
temp = temp->next = right;
right = right->next;
}
}
if( left ) temp->next = left;
else temp->next = right;
return new_list->next;
}
};效率就是最前面的runtime
當然merge函數也可以這樣寫，效率也一樣的
class Solution {
public:
ListNode* sortList(ListNode* head) {
if( !head || !head->next ) return head;
ListNode * slow = head;
ListNode * fast = head->next;
while(fast&&fast->next){
fast = fast->next->next;
slow = slow->next;
}
ListNode * second_head = slow->next;
slow->next = NULL;
return merge( sortList(head) , sortList(second_head) );
}
ListNode * merge( ListNode * left , ListNode * right ){
ListNode * new_list = new ListNode(0);
ListNode * temp = new_list;
while( left || right ){
if( left && ( !right || left->val < right->val ) ){
temp = temp->next = left;
left = left->next;
}
if( right && ( !left || left->val >= right->val ) ){
temp = temp->next = right;
right = right->next;
}
}
temp->next = NULL;
return new_list->next;
}
};至於快速排序，有點難想，就貼一個網上的寫法：
class Solution {
public:
ListNode* sortList(ListNode* head) {
ListNode* tail;
quickSort(head,tail);
return head;
}
private:
void quickSort(ListNode* &head,ListNode* &tail){
if(head==nullptr)
return;
ListNode left_head(-1),right_head(-1);
ListNode *left_tail=&left_head,*right_tail=&right_head;
ListNode *pivot_head=head,*pivot_tail=head,*now=head->next;
int key=pivot_head->val;
//paritition
while(now!=nullptr){
if(now->val < key){
left_tail->next=now;
left_tail=now;
} else if(now->val > key){
right_tail->next=now;
right_tail=now;
} else {
pivot_tail->next=now;
pivot_tail=now;
}
now=now->next;
}
//deal with end node
right_tail->next=nullptr;
left_tail->next=nullptr;
pivot_tail->next=nullptr;
quickSort(left_head.next,left_tail);
quickSort(right_head.next,right_tail);
//connect pivot,sorted left part,sorted right part
if(left_head.next!=nullptr && right_tail!=&right_head){
//left part and right part is not null
//L->P->R
left_tail->next=pivot_head;
pivot_tail->next=right_head.next;
head=left_head.next;
tail=right_tail;
} else if(left_head.next!=nullptr){
//left part is not null but right part is null
//L->P
left_tail->next=pivot_head;
head=left_head.next;
tail=pivot_tail;
} else if(right_tail!=&right_head){
//right part is not null but left part is null
//P->R
pivot_tail->next=right_head.next;
head=pivot_head;
tail=right_tail;
}
//no need to deal with two null situation.
}
};