LeetCode题解：Sort List

Sort List

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

思路：

其实可以做一个快速排序或者merge排序之类的。但是这里的linked list是单向的，所以普通的快排办不到。不过仍然可以借用快速排序的思路，每次把链表分为三个部分：小于pivot，等于pivot和大于pivot，然后对小于大于pivot的部分进行排序。最后把三个部分结合起来。这样就是一个O(nlogn)的算法了。

题解：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    struct RList
    {
        ListNode* head;
        ListNode* last;
        size_t size;
        
        RList() : head (nullptr), last (nullptr), size (0)
        {}
        
        void Append (ListNode* ln)
        {
            if (head == nullptr)
                head = last = ln;
            else
            {
                last->next = ln;
                last = last->next;
            }
            ++size;
        }
        
        void Combine (RList& l)
        {
            if (head == nullptr)
            {
                head = l.head;
                last = l.last;
            }
            else if (l.head != nullptr)
            {
                last->next = l.head;
                last = l.last;
            }
            
            size += l.size;
            
            l.size = 0;
            l.head = l.last = nullptr;
        }
    };
    
    RList sortRList(RList rl)
    {
        if (rl.size == 0)  
            return rl;
    
        int pivot_value = rl.head->val;
    
        RList lpivot, pivot, rpivot;
        
        ListNode* iter = rl.head;
        while(iter != nullptr)
        {
            if (iter->val > pivot_value)
                rpivot.Append(iter);
            else if (iter->val == pivot_value)
                pivot.Append(iter);
            else
                lpivot.Append(iter);
            iter = iter->next;
        }
    
        if (rpivot.last != nullptr) rpivot.last->next = nullptr;
        if (pivot.last != nullptr) pivot.last->next = nullptr;
        if (lpivot.last != nullptr) lpivot.last->next = nullptr;
    
        if (rpivot.size > 1) rpivot = sortRList(rpivot);
        if (lpivot.size > 1) lpivot = sortRList(lpivot);
    
        lpivot.Combine(pivot);
        lpivot.Combine(rpivot);
    
        return lpivot;
    }
    
    ListNode* sortList (ListNode* head)
    {
        if (head == nullptr)
            return nullptr;
    
        ListNode* last = head;
        int size = 1;
        while(last->next != nullptr) last = last->next, ++size;
        
        RList rl;
        rl.head = head;
        rl.last = last;
        rl.size = size;
        rl = sortRList(rl);
    
        return rl.head;
    }
};