【刷题-LeetCode】148 Sort List

148. Sort List

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

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

解法1 归并排序。用两个函数实现：

• merge：将两个有序链表合在一起
• merge_sort：将无序链表排序

找中间位置用双指针

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if(head == NULL || head->next == NULL)return head;
        ListNode *mid, *pre;
        find_mid(head, mid, pre);
        pre->next = NULL;
        return merge(sortList(head), sortList(mid));
    }
    // 找中点时，需要对mid和pre做修改，因此需要传引用
    void find_mid(ListNode* s, ListNode* &mid, ListNode* &pre){
        ListNode *pp = s;
        mid = s;
        while(pp && pp->next){
            pre = mid;
            mid = mid->next;
            pp = pp->next->next;
        }
    }
    ListNode* merge(ListNode* s1, ListNode* s2){
        ListNode *head = new ListNode, *p = s1, *q = s2;
        ListNode *cur = head;
        while(p != NULL && q != NULL){
            if(p->val < q->val){
                cur->next = p;
                p = p->next;
            }else{
                cur->next = q;
                q = q->next;
            }
            cur = cur->next;
        }
        if(p != NULL)cur->next = p;
        else cur->next = q;
        return head->next;
    }
};

解法2 快速排序。partition过程中，可以用两个链表small和large分别存储pivot左侧和右侧的数据

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if(!head || !head->next) return head;
        
        ListNode* cur = head->next;
        ListNode* small = new ListNode(0);
        ListNode* large = new ListNode(0);
        ListNode* sp = small;
        ListNode* lp = large;
        // partition
        while(cur){
            if(cur->val<head->val){
                sp->next = cur;
                sp = cur;
            }
            else{
                lp->next = cur;
                lp = cur;
            }
            cur = cur->next;
        }
        sp->next = NULL;
        lp->next = NULL;
        small=sortList(small->next);
        large=sortList(large->next);
        cur = small;
        if(cur){
            while(cur->next) cur = cur->next;
            cur->next = head;
            head->next = large;
            return small;
        }else{
            head->next = large;
            return head;
        }
    }
};