【LeetCode】23. 合并K个排序链表

链接：

https://leetcode-cn.com/problems/merge-k-sorted-lists

描述：

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

ListNode* mergeKLists(vector<ListNode*>& lists) {}

简单题

【LeetCode】21. 合并两个有序链表

https://www.cnblogs.com/crazyBlogs/p/13164038.html

思路：逐一合并

用变量 (result) 表示已经合并的链表，
循环遍历所有链表，将当前链表和 (result) 合并。

C++

展开后查看

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode* result = NULL;
        for(ListNode* list: lists){
            result = merge2Lists(result, list);
        }
        return result;
    }
};

Java

展开后查看

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode result = null;
        for(ListNode list: lists){
            result = merge2Lists(result, list);
        }
        return result;
    }
}

思路：分治合并

用分治法进行合并

C++

展开后查看

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(lists.size() == 0){
            return NULL;
        }
        return merge(lists, 0, lists.size() - 1);
    }
    ListNode* merge(vector<ListNode*>&lists, int left, int right){
        if(left == right){
            return lists[left];
        }
        int mid = left + (right - left) / 2;
        ListNode* list1 = merge(lists, left, mid);
        ListNode* list2 = merge(lists, mid + 1, right);
        return merge2Lists(list1, list2);
    }
}

Java

展开后查看

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists.length == 0){
            return null;
        }
        return merge(lists, 0, lists.length - 1);
    }
    ListNode merge(ListNode[] lists, int left, int right){
        if(left == right){
            return lists[left];
        }
        int mid = left + (right - left) / 2;
        ListNode l1 = merge(lists, left, mid);
        ListNode l2 = merge(lists, mid + 1, right);
        return merge2Lists(l1, l2);
    }
}