LeetCode:Sort List

Sort List




Total Accepted: 68684 Total Submissions: 278057 Difficulty: Medium

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

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思路:

1.nlogn的排序算法有:快排、归并、堆排;

2.可是快排对链表不适合;

3.数组中归并的空间复杂度为O(n),可是链表能够直接归并仅仅须要O(1)的空间复杂度。


code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *merge(ListNode *h1, ListNode *h2){
        if(!h1) return h2;
        if(!h2) return h1;
        
        ListNode p(0);
        ListNode *t = &p;
        while(h1 && h2){
            if(h1->val < h2->val){
                t->next = h1;
                h1 = h1->next;
            }else{
                t->next = h2;
                h2 = h2->next;
            }
            t = t->next;
        }
        
        t->next = h1 ? h1:h2;
        return p.next;
    }
    ListNode* sortList(ListNode* head) {
        
        // 空或仅仅有一个结点,直接返回
        if(!head || !head->next) return head;
        
        // 利用快、慢指针找到中间结点
        ListNode *slow = head,*fast = head;
        while(fast && fast->next && fast->next->next){
            slow = slow->next;
            fast = fast->next->next;
        }
        
        ListNode *h2 = sortList(slow->next); // 后半部分
        slow -> next = NULL; // 这里要先断链才干进行前半部分的归并
        ListNode *h1 = sortList(head); // 前半部分
        return merge(h1,h2);
    }
};


原文地址:https://www.cnblogs.com/mfmdaoyou/p/7275543.html