23 合并K个有序数组

利用21题合并两个有序数组的代码，使用for循环进行合并，效率较低；参照第一名的代码，使用分治，改变对数组的处理方法，可以大幅度提高处理效率：

修改后：

public ListNode mergeKLists(ListNode[] lists) {
		if(lists==null||lists.length==0)return null;
		return sort(lists, 0, lists.length-1);
    }
public static ListNode sort(ListNode[] lists ,int low , int high) {
	    if(low>=high) return lists[low];
	    int mid = (high - low )/2 + low;
		ListNode l1 = sort(lists ,low , mid);
		ListNode l2 = sort(lists , mid +1 ,high);
		return mergeTwoLists(l1, l2);
}

原： 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
		int length = lists.length;
		if(length==0)return null;
		if(length==1)return lists[0];
		ListNode result = lists[0];
		for(int i = 1 ;i < length; i++) {
			result = mergeTwoLists(result, lists[i]);
		}
		return result;
    }
	//重复节点值
	//default关键字修饰变量，在不同包时无法访问
	public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
		ListNode result = new ListNode(0);
		ListNode point = result;
		//{5,1 2 4}
		while(l1!=null&&l2!=null) {
				if(l1.val>l2.val) {
					point.next = new ListNode(l2.val);
					point = point.next;
					if(l2.next==null||l2.next.val>l1.val) {
						point.next = new ListNode(l1.val);
						point=point.next;
						l1=l1.next;
					}
					l2=l2.next;
				}else if(l1.vall2.val) {
						point.next = new ListNode(l2.val);
						point=point.next;
						l2=l2.next;
					}
					l1=l1.next;
				}else {
					point.next = new ListNode(l2.val);
					point=point.next;
					point.next = new ListNode(l2.val);
					point=point.next;
					l1=l1.next;
					l2=l2.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 result.next;
	}
    
}