23. 合并K个排序链表

2018-07-16 02:39:27来源：博客园阅读 ()

知乎ID: 码蹄疾
码蹄疾，毕业于哈尔滨工业大学。
小米广告第三代广告引擎的设计者、开发者；
负责小米应用商店、日历、开屏广告业务线研发；
主导小米广告引擎多个模块重构；
关注推荐、搜索、广告领域相关知识;

题目

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。
示例:
输入:
[
1->4->5,
1->3->4,
2->6
]
输出: 1->1->2->3->4->4->5->6

分析

前面已经做过两个有序链表的合并，只要采用二分，分而治之，两两合并即可。时间复杂度方面，合并两个链表的长度的时间复杂度是o(min(m, n))，其中m，n分别是链表的长度。二合并的长度是o(logk)的时间复杂度。所以整体的时间复杂度为o(klogn)

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        ListNode merged = null;
        ListNode head = null;
        while (l1 != null && l2 != null) {
            if (head == null) {
                if (l1.val < l2.val) {
                    merged = l1;
                    l1 = l1.next;
                } else {
                    merged = l2;
                    l2 = l2.next;
                }
                head = merged;
                continue;
            }

            if (l1.val < l2.val) {
                merged.next = l1;
                l1 = l1.next;
            } else {
                merged.next = l2;
                l2 = l2.next;
            }
            merged = merged.next;
        }

        while (l1 != null) {
            merged.next = l1;
            l1 = l1.next;
            merged = merged.next;
        }

        while (l2 != null) {
            merged.next = l2;
            l2 = l2.next;
            merged = merged.next;
        }
        return head;
    }

    public ListNode mergeHelper(ListNode[] lists, int low, int high) {
        if (low < high) {
            int mid = (low + high) / 2;
            ListNode leftList = mergeHelper(lists, low, mid);
            ListNode rightList = mergeHelper(lists, mid + 1, high);
            return mergeTwoLists(leftList, rightList);
        }
        return lists[low];
    }


    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        return mergeHelper(lists, 0, lists.length - 1);
    }
}

拓展

合并两个有序链表，之前采用的是非递归的解法。感觉代码有点长，可以采用递归的解法，缩短代码量。合并的时候选最小的元素，链表头指针后移动，递归合并即可。

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) {
            return null;
        }

        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        ListNode merged;
        if (l1.val > l2.val) {
            merged = l2;
            l2 = l2.next;
            merged.next = mergeTwoLists(l1, l2);
        } else {
            merged = l1;
            l1 = l1.next;
            merged.next = mergeTwoLists(l1, l2);
        }
        return merged;
    }
}