链表算法题一

旋转链表

在这里插入图片描述

首先考虑特殊情况

若给定链表为空表或者单个节点,则直接返回head,不需要旋转操作.
题目给定条件范围: $0 <= k <= 2 * 10^9$ ,但是受给定链表长度的限制,比如示例2中,k=4与k=1的效果等价.
那么可以得出k=k%len的式子,其中len为数组长度.
先遍历原链表,求得len.
求得新链表的头节点,尾节点在原链表中位置,修改指针指向即可.

class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head==null||head.next==null) return head;
        int len = 0;
        //遍历求链表长度并且求出原链表的末尾节点.
        ListNode tail = head;
        while(tail.next!=null)
        {
            tail = tail.next;
            len++;
        }
        len++;
        //处理k
        k = k % len;
        if(k==0) return head;
        //找新链表的尾节点.
        int n = len-k-1;
        ListNode cur = head;
        while(n>0)
        {
            cur = cur.next;
            n--;
        }
        tail.next = head;
        //找到新链表的头节点,其后修改指针指向即可.
        head = cur.next;
        cur.next = null;
        return head;
    }
}

合并K个链表

分治思想+归并排序

注意此题与数组的归并排序区别.
分治部分和数组相同,但合并部分merge函数实际是此题:合并两个有序链表.
如果了解归并排序和做个上面那道题,思路一通水到渠成.
结论:链表的归并排序空间复杂度: $O (1)$

class Solution {
    private ListNode mergeListSort(ListNode[] lists,int start,int end)
    {
            if(start>end)
                return null;
            if(start==end)
                return lists[start];
            int mid = start + (end-start)/2;
            ListNode left = mergeListSort(lists,start,mid);
            ListNode right = mergeListSort(lists,mid+1,end);
            return merge(left,right);
    }
    private ListNode merge(ListNode left,ListNode right)
    {
            if(left==null)
                return right;
            if(right==null)
                return left;
            ListNode cur1 = left,cur2 = right;
            ListNode head = new ListNode();
            ListNode tail = head;
            while(cur1!=null&&cur2!=null)
            {
                if(cur1.val<=cur2.val)
                {
                    tail.next = cur1;
                    cur1 = cur1.next;
                    tail = tail.next;
                }
                else
                {
                    tail.next = cur2;
                    cur2 = cur2.next;
                    tail = tail.next;
                }
            }
            if(cur1!=null)
                tail.next=cur1;
            if(cur2!=null)
                tail.next=cur2;
            return head.next;
    } 
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists==null||lists.length==0)
            return null;
        //MergeSort启动!
        return mergeListSort(lists,0,lists.length-1);
        //当lists.length==1时,上式会返回lists.
    }
}

力扣只写了两道题的笔记,太累了写不动ε(┬┬﹏┬┬)3.
力扣折磨.