Sort a linked list in O(n log n) time using constant space complexity.
/*
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
/
public class Solution {
public ListNode sortList(ListNode head) {
}
}
1
一看到O(nlogn)就会想到排序算法，刚好归并排序适用于数组和链表，时间复杂度是O(nlogn)。
注意findMid()这个方法，这个是O(n)时间的查找链表中间元素的方法，使用两个指针遍历链表，
使两个指针之间有某种关系来解决问题，寻找链表中倒数第k个元素也是这种方法。
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public ListNode mergeSort(ListNode head){
if(head == null || head.next == null) return head;
ListNode mid = findMid(head);
ListNode r = mid.next;
mid.next = null;
ListNode left = mergeSort(head);
ListNode right = mergeSort(r);
return merge(left,right);
}
private ListNode merge(ListNode left,ListNode right){
ListNode dummyHead = new ListNode(0);
ListNode cur = dummyHead;
while(left != null && right != null){
if(left.val < right.val){
cur.next = left;
left = left.next;
}else{
cur.next = right;
right = right.next;
}
cur = cur.next;
}
cur.next = (left == null) ? right : left;
return dummyHead.next;
}
private ListNode findMid(ListNode head){
ListNode slow,fast;
slow = fast = head;
while(fast.next != null && fast.next.next != null){
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
2 leetcode discuss
most votes的方法也是这个。