leetcode之链表排序

147. 对链表进行插入排序

class Solution {
    
    
    public ListNode insertionSortList(ListNode head) {
    
    
        if(head == null || head.next == null){
    
    
            return head;
        }
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode last = head; // 指向链表已排序部分的最后一个结点
        ListNode cur = head.next; // 指向待插入的元素
        while(cur != null){
    
    
            if(last.val <= cur.val){
    
    
                last = last.next;
            }else{
    
    
                ListNode pre = dummyHead; // 指向 cur 要插入位置的前一个位置
                while(pre.next != null && pre.next.val < cur.val){
    
    
                    pre = pre.next;
                }
                last.next = cur.next;
                cur.next = pre.next;
                pre.next = cur;
            }
            cur = last.next;
        }
        return dummyHead.next;
    }
}

148. 排序链表

class Solution {
    
    
    public ListNode sortList(ListNode head) {
    
    
        if(head == null || head.next == null){
    
    
            return head;
        }
        ListNode fast = head, slow = head;
        // 注意：不能是 while(fast != null && fast.next != null)，否则会无限递归。比如当链表为：4->2->null 时，就会出错。
        while(fast.next != null && fast.next.next != null){
    
    
            fast = fast.next.next;
            slow = slow.next;
        }
        ListNode node = slow.next;
        slow.next = null;
        ListNode left = sortList(head);
        ListNode right = sortList(node);
        return merge(left, right);
    }
    public 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;
        }
        if(left != null){
    
    
            cur.next = left;
        }
        if(right != null){
    
    
            cur.next = right;
        }
        return dummyHead.next;
    }
}

• 时间复杂度 $O(nlogn)$，空间复杂度 $O(logn)$