• [LeetCode] Sort List


    There are many merge-sort solutions at the forum, but very few quicksort solutions. So I post my accepted quicksort solution here.

    Well, after reading the problem statement, I intuitively select quicksort since it is able to give an in-place solution and thus costs only constant space. Also, it is O(nlogn) in the expected case though it may become O(n^2) in the worst case.

    Then I implement my quicksort solution and test it. I then submit it to the online judge. However, the annoying TLE error occurred. I check for the forums and some people suggested to use random pivoting or duplicate skipping. However, implementing random pivoting is a little costly, I lazily tried to skip the duplicates. And it works! So now comes the following solution . Note that each time I choose the first node as the pivot. Moreover, I create a new_head that points to headfor convenience.

    Of course, this solution passes the online judge luckily. If the linked list is like: 100000 -> 99999 -> 99998 -> ... -> 1, it will fail since the subproblems only decrease by 1 at each recursion. However, it seems that the LeetCode OJ does not have this kind of test cases.

     1     void sortListHelper(ListNode* head, ListNode* tail) {
     2         if (head -> next == tail) return;
     3         /* Partition the list. */
     4         ListNode* pre = head;
     5         ListNode* cur = head -> next;
     6         ListNode* pivot = cur;
     7         while (cur -> next && cur -> next != tail) {        
     8             if (pivot -> val > cur -> next -> val) {
     9                 ListNode* temp = pre -> next;
    10                 pre -> next = cur -> next;
    11                 cur -> next = cur -> next -> next;
    12                 pre -> next -> next = temp;
    13             }
    14             else cur = cur -> next;
    15         }
    16         sortListHelper(head, pivot);
    17         /* Here is the trick. */
    18         while (pivot -> next != tail && pivot -> next -> val == pivot -> val)
    19             pivot = pivot -> next;
    20         if (pivot -> next != tail) sortListHelper(pivot, tail);
    21     } 
    22 
    23     ListNode* sortList(ListNode* head) {
    24         ListNode* new_head = new ListNode(0);
    25         new_head -> next = head;
    26         sortListHelper(new_head, NULL);
    27         return new_head -> next;
    28     }
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  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4548064.html
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