LeetCode 148. Sort List
原题链接在这里：https://leetcode.com/problems/sort-list/
题目：
Sort a linked list in O(n log n) time using constant space complexity.
题解：
divide and conquer. 从中点开始把原list段成两个list. 一直拆到只有一个点，再按照Merge Two Sorted Lists给合并回来。
Time Complexity: O(nlogn). Space: O(logn), 用了logn层stack.
AC Java:
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode sortList(ListNode head) { 11 if(head == null || head.next == null){ 12 return head; 13 } 14 ListNode mid = findMiddle(head); //找到中点 15 ListNode secondHead = mid.next; //中点后一个点最为另一个head. 16 mid.next = null; //断开 17 return mergeTwoSortedLists(sortList(head), sortList(secondHead)); //divide and conquer 的merge sort 18 } 19 private ListNode findMiddle(ListNode head){ 20 ListNode walker = head; 21 ListNode runner = head; 22 while(runner.next != null && runner.next.next != null){ 23 walker = walker.next; 24 runner = runner.next.next; 25 } 26 return walker; 27 } 28 private ListNode mergeTwoSortedLists(ListNode l1, ListNode l2){ 29 if(l1 == null){ 30 return l2; 31 } 32 if(l2 == null){ 33 return l1; 34 } 35 ListNode dummy = new ListNode(0); 36 ListNode cur = dummy; 37 while(l1 != null && l2 != null){ 38 if(l1.val <= l2.val){ 39 cur.next = l1; 40 cur = cur.next; 41 l1 = l1.next; 42 }else{ 43 cur.next = l2; 44 cur = cur.next; 45 l2 = l2.next; 46 } 47 } 48 if(l1 != null){ 49 cur.next = l1; 50 } 51 if(l2 != null){ 52 cur.next = l2; 53 } 54 return dummy.next; 55 } 56 }
AC C++:
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode() : val(0), next(nullptr) {} 7 * ListNode(int x) : val(x), next(nullptr) {} 8 * ListNode(int x, ListNode *next) : val(x), next(next) {} 9 * }; 10 */ 11 class Solution { 12 public: 13 ListNode* sortList(ListNode* head) { 14 if(!head || !head->next){ 15 return head; 16 } 17 18 ListNode* mid = findMiddle(head); 19 ListNode* midNext = mid->next; 20 mid->next = nullptr; 21 return mergeTwoSorted(sortList(head), sortList(midNext)); 22 } 23 24 ListNode* findMiddle(ListNode* head){ 25 ListNode* walker = head; 26 ListNode* runner = head; 27 while(runner && runner->next && runner->next->next){ 28 walker = walker->next; 29 runner = runner->next->next; 30 } 31 32 return walker; 33 } 34 35 ListNode* mergeTwoSorted(ListNode* l1, ListNode* l2){ 36 if(!l1){ 37 return l2; 38 } 39 40 if(!l2){ 41 return l1; 42 } 43 44 ListNode* dummy = new ListNode(0); 45 ListNode* it = dummy; 46 while(l1 && l2){ 47 if(l1->val <= l2->val){ 48 it->next = l1; 49 l1 = l1->next; 50 }else{ 51 it->next = l2; 52 l2 = l2->next; 53 } 54 55 it = it->next; 56 } 57 58 if(l1){ 59 it->next = l1; 60 } 61 62 if(l2){ 63 it->next = l2; 64 } 65 66 return dummy->next; 67 } 68 };
