LeetCode —— Sort List

做了一段时间的leetcode，别看leetcode的算法基础，边界条件cover的非常全面，

题目要求时间复杂度O(n log n)，空间复杂度为常量，首先想到快速排序，堆排序，但这两种算法比较适用于数组，

就决定用非常基本的并归排序算法，通过调整node的位置，进行二路合并

思路虽然简单，遇到相同 val的节点，一不小心会丢失，终于找到了原因，分析并通过，总结一下，帮助提高。

public class Solution {    public ListNode sortList(ListNode head) {        if(head==null||head.next ==null)return head;            ListNode ans = null;        ListNode slow = null,fast = null;        for(slow = head,fast = head;fast.next != null&&fast.next.next != null;        slow = slow.next,fast = fast.next.next);        ListNode head2 = slow.next;        slow.next = null;        if(head != slow){        ans = merge(head,head2);        }else{        if(head.val>head2.val){        ans = head2;        head2.next = head;        }else{        head.next = head2;        ans = head;        }        }        return ans;    }    ListNode merge(ListNode head1,ListNode head2){    ListNode slow = null,fast = null;    if(head1.next == null){    //*************one node in list1, do nothing**********    }else if(head1.next.next == null){    //*************two node in list1, judge and adjust*********    if(head1.val > head1.next.val){    head1.next.next = head1;    head1 = head1.next;    head1.next.next = null;    }    }else{        //***************more than two node in list1, recursion*********    for(slow = head1,fast = head1;fast.next != null&&fast.next.next != null;    slow = slow.next,fast = fast.next.next);    ListNode tmp = slow.next;    slow.next = null;    head1 = merge(head1,tmp);    }        if(head2.next == null){    //*************one node in list2, do nothing**********    }else if(head2.next.next==null){    //*************two node in list2, judge and adjust*********    if(head2.val > head2.next.val){    head2.next.next = head2;    head2 = head2.next;    head2.next.next=null;    }    }else{        //***************more than two node in list2, recursion*********    for(slow = head2,fast = head2;fast.next != null&&fast.next.next != null;    slow = slow.next,fast = fast.next.next);    ListNode tmp = slow.next;    slow.next = null;    head2 = merge(head2,tmp);    }        ListNode ans = null;    if(head1.val > head2.val) ans = head2;    else ans = head1;        while(head1!=null && head2!=null){    ListNode tmp = null;    /****                * 在合并阶段，非常关键，小心丢失相同元素                * head1.val == head2.val 时，head1向前推进;                * 所以当head1.next.val == head2.val, 不能改变head1.next,直接后移                * 当head2.next.val == head1.val, 要改变head2.next指向head1，head2再向后移。                */    if(head1.val > head2.val) {     tmp = head2.next;    if(tmp!=null){    if(tmp.val < head1.val){ //这里不能有  =    head2 = head2.next;    }else{    head2.next = head1;    head2 = tmp;    }    }else{    head2.next = head1;    head2 = null;    }    }else{    tmp = head1.next;    if(tmp!=null){    if(tmp.val <= head2.val){//head1.next.val == head2.val 时 ， head1后移但不改变head1.next指向head2    head1 = head1.next;  //这里要有 =    }else{    head1.next = head2;    head1 = tmp;    }    }else{    head1.next = head2;    head1 = null;    }    }    }    return ans;    }    }