23.合并K个升序链表¶

题目描述¶

原题

给你一个链表数组，每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中，返回合并后的链表。

示例 1：

输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

示例 2：

1 2	`输入：lists = [] 输出：[]`

示例 3：

1 2	`输入：lists = [[]] 输出：[]`

提示：

k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按升序排列
lists[i].length 的总和不超过 10^4

题解¶

解法一¶

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
//执行用时: 278 ms
//内存消耗: 40.9 MB
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists == null || lists.length == 0)return null;
        ListNode cur = lists[0];
        //遍历所有的链表
        for(int i = 1;i < lists.length;i++){
            cur = merge(cur,lists[i]);
        }
        return cur;
    }
    //两个有序链表合并
    private ListNode merge(ListNode l1,ListNode l2){
        if(l1==null)return l2;
        if(l2==null)return l1;
        if(l1.val <= l2.val){
            l1.next = merge(l1.next,l2);
            return l1;
        }else{
            l2.next = merge(l1,l2.next);
            return l2;
        }
    }
}

解法二分治算法¶

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
//
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists == null || lists.length == 0) return null;
        sort(lists,0,lists.length-1);
        return lists[0];
    }
    //参考算法第四版的归并排序
    //1. 利用sort方法拆分成[1] [2] [3] [4]
    //2. 归并
    //和归并排序的区别
    //merge的时候不需要知道开始和结束的索引
    private void sort(ListNode[] lists, int lo, int hi) {
        if (hi <= lo) return;
        int mid = lo + (hi - lo) / 2;
        sort(lists, lo, mid); //将左半边排序
        sort(lists, mid + 1, hi); //将右半边排序
        lists[lo] = merge(lists[lo],lists[mid+1]); //归并结果
        lists[mid+1]=null;
    }
    private ListNode merge(ListNode l1,ListNode l2){
        if(l1==null)return l2;
        if(l2==null)return l1;
        if(l1.val <= l2.val){
            l1.next = merge(l1.next,l2);
            return l1;
        }else{
            l2.next = merge(l1,l2.next);
            return l2;
        }
    }
}

//官方分治算法代码
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return merge(lists, 0, lists.length - 1);
    }

    public ListNode merge(ListNode[] lists, int l, int r) {
        if (l == r) {
            return lists[l];
        }
        if (l > r) {
            return null;
        }
        int mid = (l + r) >> 1;
        return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));
    }

    public ListNode mergeTwoLists(ListNode a, ListNode b) {
        if (a == null || b == null) {
            return a != null ? a : b;
        }
        ListNode head = new ListNode(0);
        ListNode tail = head, aPtr = a, bPtr = b;
        while (aPtr != null && bPtr != null) {
            if (aPtr.val < bPtr.val) {
                tail.next = aPtr;
                aPtr = aPtr.next;
            } else {
                tail.next = bPtr;
                bPtr = bPtr.next;
            }
            tail = tail.next;
        }
        tail.next = (aPtr != null ? aPtr : bPtr);
        return head.next;
    }
}