【LeetCode】Sort List

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题目

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

解答

O(nlogn)时间复杂度的排序有快排、堆排、归并，一般双向链表用快排、单向链表用归并，堆排两种都可以，以下使用归并排序：

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } *///归并排序public class Solution {    public ListNode sortList(ListNode head) {        if(head==null||head.next==null){        return head;        }else{        //快慢指针找到中间点        ListNode fast=head;        ListNode slow=head;        while(fast.next!=null&&fast.next.next!=null){        fast=fast.next.next;        slow=slow.next;        }        fast=slow;             //注意修改fast和slow的值，若当其为一个节点时，就不会调用sortList        slow=slow.next;        fast.next=null;        fast=sortList(head);  //前半段排序        slow=sortList(slow);  //后半段排序        return mergeArray(fast,slow);        }    }    ListNode mergeArray(ListNode list1,ListNode list2){    if(list1==null){    return list2;    }    if(list2==null){    return list1;    }    ListNode mergeList=null;    if(list1!=null&&list2!=null){    if(list1.val<list.val){    mergeList=list1;    list1=list1.next;    mergeList.next=null;    }else{    mergeList=list2;    list2=list2.next;    mergeList.next=null;    }    }    ListNode tempList=mergeList;    while(list1!=null&&list2!=null){    if(list1.val<list2.val){    tempList.next=list1;    list1=list1.next;    tempList=tempList.next;    tempList.next=null;    }else{    tempList.next=list2;    list2=list2.next;    tempList=tempList.next;    tempList.next=null;    }    }    if(list1!=null){    tempList.next=list1;    }    if(list2!=null){    tempList.next=list2;    }    return mergeList;    }}

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