leetcode_SortList

描述:

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

在O(n*log(n))的时间复杂度,常数级空间复杂度内对一个链表进行排序
采用归并排序,用快慢指针将链表分成两部分,最后合并两个链表.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* sortList(ListNode* head) {        if(!head || !head->next) return head;        ListNode * p = head;        ListNode * q = head->next;        //二分        while(q && q->next){            p = p->next;            q = q->next->next;        }        ListNode * right = sortList(p->next);    //对后半段递归排序        p->next = NULL; //把前后两段分开        ListNode * left = sortList(head);        //合并前后两段,实际上是用这里的判断大小来排序        return merge(left, right);    }    // 合并函数    ListNode *merge(ListNode *l, ListNode *r){        ListNode res(0);        ListNode * p = &res;        //按增序合并l和r        while(l && r){            if(l->val < r->val){                p->next = l;                l = l->next;            }else{                p->next = r;                r = r->next;            }            p = p->next;        }        if(l) p->next = l;        if(r) p->next = r;        return res.next;    }};