23. Merge k Sorted Lists

solution1 - use divide and conquer

divide and conquer => logk, k is lists[] length, split to 2 part

merge O(N), N is all nodes in two lists

, so O(Nlogk)

notice, in this case, should return null => . output: []

/**
 * 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; // notice this!
        
        return partition(lists, 0, lists.length - 1);
    }
    
    private ListNode partition(ListNode[] lists, int s, int e) {
        if (s == e) {
            return lists[s]; // notice this!
        } else if (s < e) {
            int mid = (s+e)/2;
            ListNode l1 = partition(lists, s, mid);
            ListNode l2 = partition(lists, mid+1, e);
            return merge(l1, l2);
        } else {
            return null; // notice this!
        }
    }
    
    // this part is leetcode 21. Merge Two Sorted Lists
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(-1); // notice this should new a node
        ListNode pre = head;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                pre.next = l1;
                l1 = l1.next;
            } else {
                pre.next = l2;
                l2 = l2.next;
            }
            pre = pre.next;
        }
        pre.next = (l1 == null) ? l2 : l1;
        
        return head.next;
    }
}

假設只有兩個數, index 0, 1

所以 mid = 0+1)/2 = 0

l1 = partition(lists, 0,0) = ( s == e return lists[s] ) = list[0]

l2 = partition(lists, 0+1,1) = ( s == e return lists[s] ) = list[1]

return merge(list[0], list[1])

    private ListNode partition(ListNode[] lists, int s, int e) {
        if (s == e) {
            return lists[s]; // notice this!
        } else if (s < e) {
            int mid = (s+e)/2;
            ListNode l1 = partition(lists, s, mid);
            ListNode l2 = partition(lists, mid+1, e);
            return merge(l1, l2);
        } else {
            return null; // notice this!
        }
    }

solution2 - use priorityQueue

easier solution!

/**
 * 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;
        PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, (a, b) -> a.val - b.val);
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        
        for (ListNode list : lists) {
            if (list != null) {
                queue.add(list);
            }
        }
        while (!queue.isEmpty()) {
            cur.next = queue.poll();
            cur = cur.next;
            if (cur.next != null) {
                queue.add(cur.next);
            }
        }
        return dummy.next;
    }
}