Merge k Sorted Lists

lc23

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

version 1: Divide & Conquer

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param lists: a list of ListNode
     * @return: The head of one sorted list.
     */
    public ListNode mergeKLists(List<ListNode> lists) {
        if (lists.size() == 0) {
            return null;
        }
        return mergeHelper(lists, 0, lists.size() - 1);
    }

    private ListNode mergeHelper(List<ListNode> lists, int start, int end) {
        if (start == end) {
            return lists.get(start);
        }

        int mid = start + (end - start) / 2;
        ListNode left = mergeHelper(lists, start, mid);
        ListNode right = mergeHelper(lists, mid + 1, end);
        return mergeTwoLists(left, right);
    }

    private ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                tail.next = list1;
                tail = tail.next;
                list1 = list1.next;
            } else {
                tail.next = list2;
                tail = tail.next;
                list2 = list2.next;
            }
        }
        if (list1 != null) {
            tail.next = list1;
        } else {
            tail.next = list2;
        }

        return dummy.next;
    }
}

version 2: Heap

public class Solution {

       private Comparator<ListNode> ListNodeComparator = new Comparator<ListNode>() {
        public int compare(ListNode left, ListNode right) {
            if (left == null) {
                return 1;
            } else if (right == null) {
                return -1;
            }
            return left.val - right.val;
        }
    };

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


        //注意九章这个写法
       //        Queue<ListNode> heap = new PriorityQueue<ListNode>(lists.size(), ListNodeComparator);
        Queue<ListNode> heap = new PriorityQueue<ListNode>(lists.length, new Comparator<ListNode>(){

            public int compare(ListNode l1, ListNode l2){
                if(l1 == null) return -1;
                if(l2 == null) return 1;
                return l1.val - l2.val;
            }


        });
        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null) {
                heap.add(lists[i]);
            }
        }

        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while (!heap.isEmpty()) {
            ListNode head = heap.poll();
            tail.next = head;
            tail = tail.next;
            if (head.next != null) {
                heap.add(head.next);
            }
        }
        return dummy.next;
    }
}

Version 3: merge two by two

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param lists: a list of ListNode
     * @return: The head of one sorted list.
     */
    public ListNode mergeKLists(List<ListNode> lists) {  
        if (lists == null || lists.size() == 0) {
            return null;
        }

        while (lists.size() > 1) {
            List<ListNode> new_lists = new ArrayList<ListNode>();
            for (int i = 0; i + 1 < lists.size(); i += 2) {
                ListNode merged_list = merge(lists.get(i), lists.get(i+1));
                new_lists.add(merged_list);
            }
            if (lists.size() % 2 == 1) {
                new_lists.add(lists.get(lists.size() - 1));
            }
            lists = new_lists;
        }

        return lists.get(0);
    }

    private ListNode merge(ListNode a, ListNode b) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while (a != null && b != null) {
            if (a.val < b.val) {
                tail.next = a;
                a = a.next;
            } else {
                tail.next = b;
                b = b.next;
            }
            tail = tail.next;
        }

        if (a != null) {
            tail.next = a;
        } else {
            tail.next = b;
        }

        return dummy.next;
    }
}