Although, sorting a linked list can be done via Recursive Divide-and-Conquer algorithm i.e. merge sorting, we can however, turn the linked list into an array (or vector) using O(N) time and space, then sort the array/vector in O(nlogn), and finally convert it back to the linked list in O(n) time.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* insertionSortList(ListNode* head) { if (head == NULL) return NULL; vector<int> data; ListNode *p = head; while (p) { data.push_back(p->val); p = p->next; } sort(begin(data), end(data)); p = head; for (const auto &n: data) { p->val = n; p = p->next; } return head; } }; |
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* insertionSortList(ListNode* head) {
if (head == NULL) return NULL;
vector<int> data;
ListNode *p = head;
while (p) {
data.push_back(p->val);
p = p->next;
}
sort(begin(data), end(data));
p = head;
for (const auto &n: data) {
p->val = n;
p = p->next;
}
return head;
}
};We don’t need to allocate new nodes for the sorted singly-linked list. Instead, we can follow the original linked list in the same order of the sorted array, then synchronise the values from the array to the linked list. This will cost O(N) time and O(1) additional space.
–EOF (The Ultimate Computing & Technology Blog) —
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