【LeetCode】Merge k Sorted Lists

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Merge k Sorted Lists
Total Accepted: 13489 Total Submissions: 61228 My Submissions
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
【解题思路】
1、参考Merge Two Sorted Lists，依次归并，每个节点都至少需要扫描一次，很耗时间。
2、参考地址LeetCode:Merge k Sorted Lists，分析的很详细。
这里采取了第二种方法。后续有时间会更新内容。

Java AC 516ms

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode mergeKLists(ArrayList<ListNode> lists) {if (lists == null || lists.size() == 0) {return null;}int n = lists.size();while (n > 1) {int k = (n + 1) / 2;for (int i = 0; i < n / 2; i++) {lists.set(i, mergeTwoLists(lists.get(i), lists.get(i + k)));}n = k;}return lists.get(0);}public ListNode mergeTwoLists(ListNode l1, ListNode l2) {ListNode node = new ListNode(0);ListNode point = node;while (l1 != null && l2 != null) {if (l1.val > l2.val) {point.next = new ListNode(l2.val);point = point.next;l2 = l2.next;} else {point.next = new ListNode(l1.val);point = point.next;l1 = l1.next;}}while (l1 != null) {point.next = new ListNode(l1.val);point = point.next;l1 = l1.next;}while (l2 != null) {point.next = new ListNode(l2.val);point = point.next;l2 = l2.next;}return node.next;}}

Python AC 1116ms

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = None class Solution:    # @param a list of ListNode    # @return a ListNode    def mergeKLists(self, lists):        if lists is None or len(lists) == 0:            return None        n = len(lists)        while n > 1:            k = (n + 1) / 2            for i in range(n/2):                lists[i] = self.mergeTwoLists(lists[i], lists[i+k])            n = k        return lists[0]    def mergeTwoLists(self, l1, l2):        node = ListNode(0)        point = node        while l1 is not None and l2 is not None:            if l1.val > l2.val:                point.next = ListNode(l2.val)                l2 = l2.next            else:                point.next = ListNode(l1.val)                l1 = l1.next            point = point.next        while l1 is not None:            point.next = ListNode(l1.val)            point = point.next            l1 = l1.next        while l2 is not None:            point.next = ListNode(l2.val)            point = point.next            l2 = l2.next        return node.next

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