Sort List

来源：互联网发布：乐乎城市青年社区北苑编辑：程序博客网时间：2024/06/10 08:32

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

用归并算法，先找到 list 的中点，然后用 divide and conquer

1. 用 slow,fast 找到中点

2. 合并两个 sorted list

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param head, a ListNode    # @return a ListNode    def sortList(self, head):        return self.mergesort(head)            def mergesort(self,head):        if head==None or head.next==None:            return head                slow = fast = head        while(fast.next!=None and fast.next.next!=None):            slow = slow.next            fast = fast.next.next        sec = slow.next        slow.next = None                head = self.mergesort(head)        sec = self.mergesort(sec)        return self.mergetwo(head,sec)        def mergetwo(self,first,sec):        dummy = ListNode(100)        cur = dummy        cur1,cur2 = first, sec                while True:            if cur1==None:                cur.next = cur2                return dummy.next            if cur2==None:                cur.next = cur1                return dummy.next                        if cur1.val < cur2.val:                cur.next = cur1                cur = cur.next                cur1 = cur1.next            else:                cur.next = cur2                cur = cur.next                cur2 = cur2.next

c++

/** * 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) {        mergesort(head);          }        ListNode *mergesort(ListNode *head){        if(head==NULL || head->next==NULL){            return head;        }                ListNode *slow = head, *fast = head;        while(fast->next!=NULL && fast->next->next!=NULL){            slow = slow->next;            fast = fast->next->next;        }        ListNode *sec = slow->next;        slow->next = NULL;        head = mergesort(head);        sec = mergesort(sec);        return mergetwo(head,sec);    }        ListNode *mergetwo(ListNode *first, ListNode *sec){        ListNode *dummy = new ListNode(100);        ListNode *cur=dummy, *cur1=first, *cur2=sec;        while(1){            if(cur1==NULL){                cur->next = cur2;                return dummy->next;            }            if(cur2==NULL){                cur->next = cur1;                return dummy->next;            }            if(cur1->val<cur2->val){                cur->next = cur1;                cur1 = cur1->next;            }else{                cur->next = cur2;                cur2 = cur2->next;            }            cur = cur->next;        }    }    };

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