知乎ID: 碼蹄疾
碼蹄疾，畢業於哈爾濱工業大學。
小米廣告第三代廣告引擎的設計者、開發者；
負責小米應用商店、日歷、開屏廣告業務線研發；
主導小米廣告引擎多個模塊重構；
關註推薦、搜索、廣告領域相關知識;

題目

合並 k 個排序鏈表，返回合並後的排序鏈表。請分析和描述算法的復雜度。
示例:
輸入:
[
1->4->5,
1->3->4,
2->6
]
輸出: 1->1->2->3->4->4->5->6

分析

前面已經做過兩個有序鏈表的合並，只要采用二分，分而治之，兩兩合並即可。時間復雜度方面，合並兩個鏈表的長度的時間復雜度是o(min(m, n))，其中m，n分別是鏈表的長度。二合並的長度是o(logk)的時間復雜度。所以整體的時間復雜度為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

合並兩個有序鏈表，之前采用的是非遞歸的解法。感覺代碼有點長，可以采用遞歸的解法，縮短代碼量。合並的時候選最小的元素，鏈表頭指針後移動，遞歸合並即可。

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;
    }
}

相關題目

21. 合並兩個有序鏈表

23. 合並K個排序鏈表