[LeetCode] 23. Merge k Sorted Li

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

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Solution

This is the advanced version of Merge Two Sorted Lists.
Merge Two Sorted Lists

We can use the function of Merge Two Sorted Lists; and by using recursion, we can achieve the goal.

However, we should not deal with two strings over and over again like code shown below.

for(int i=1;i<lists.length;i++){
    answerNode = mergeTwoLists(answerNode,lists[i]);
}

This code will lead to TLE, instead we should divide the problem and first merge two different lists, and then merge two merged lists, and over and over. In this way, we can reduce the usage of MergeTwoLists function from O(n) to O(O/2+O/4+...).

The code is shown as below.

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        
        if (lists == null || lists.length==0)  return null;
        
        return Partion(lists,0,lists.length-1);
    }
    
    public ListNode Partion(ListNode[] lists, int start, int end){
        
        //Output condition
        if(start == end)
            return lists[start];
            
        if(start<end){
            int mid = (start+end)/2;
            ListNode l1 = Partion(lists,start,mid);
            ListNode l2 = Partion(lists,mid+1,end);
            return mergeTwoLists(l1,l2);
        }
        else
            return null;
    }
    
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        
        ListNode mergeHead;
        
        if (l1 == null)
            return l2;
        if(l2 == null)
            return l1;
        
        if(l1.val<l2.val){
            mergeHead = l1;
            mergeHead.next = mergeTwoLists(l1.next,l2);
        }
        else {
            mergeHead = l2;
            mergeHead.next = mergeTwoLists(l1,l2.next);
        }
        return mergeHead;
    }
}

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