Leetcode | sort list

Sort a linked list in O (n log n) time using constant space complexity.

Merge sort, Heap Sort, and quick sort are both O (nlgn), but mergesort and quicksort are recursive, rather than constant space. heapsort requires random access.

The bottom up implementation of mergesort can be constant space. It is also applicable to linked lists.

If it is an array, you can also use heapsort.

 1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     // merge two list, and return last element (not NULL)12     ListNode* merge(ListNode *nh, ListNode *p1, ListNode *p2) {13         while (p1 || p2) {14             if (p1 && (p2 == NULL || p1->val < p2->val)) {15                 nh->next = p1;16                 p1 = p1->next;17             } else {18                 nh->next = p2;19                 p2 = p2->next;20             }21             nh = nh->next;22         }23         return nh;24     }25     26     // get length of list27     int getLength(ListNode *head) {28         int n = 0;29         while (head) {30             n++;31             head = head->next;32         }33         return n;34     }35     36     ListNode *sortList(ListNode *head) {37         int n = getLength(head);38         ListNode *p1, *p2, *tmp, *newH1, *tail;39         40         // merge sort, bottom up41         for (int l = 1; l < n; l *= 2) {42             p1 = head;43             tail = head = NULL; // head of the whole list44             while (p1) {45                 p2 = p1;46                 for (int i = 1; i < l && p2; ++i) {47                     p2 = p2->next;48                 }49                 if (!p2) break;50                 tmp = p2->next;51                 p2->next = NULL; // set tail of list 1 to NULL52                 p2 = tmp; 53                 for (int i = 1; i < l && tmp; ++i) {54                     tmp = tmp->next;55                 }56                 if (tmp) {57                     newH1 = tmp->next; // get next head of list 158                     tmp->next = NULL; // set tail of list 2 to NULL59                 } else {60                     newH1 = NULL;61                 }62                 ListNode h(0);63                 ListNode *last = merge(&h, p1, p2);64                 if (tail) tail->next = h.next; // connect the sorted part with the current two list65                 if (!head) head = h.next;66                 tail = last;67                 last->next = newH1;68                 p1 = newH1;69             }70         }71         return head;72     }73 };