23. Merge k Sorted Lists

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

 

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

 

// 常规做法是正常的两两合并，最终将多个合并成一个。即是先将前两个合并，然后再与第三个合并，再与第四个，第五个。。。直到所有合并结束

// 使用这种二分思想，可以减少遍历的节点数量

public ListNode mergeKLists(ListNode[] lists) {
    if(lists.length==0) return null;
    // 从外向内合并首尾两个链表
    int rightIndex = lists.length-1;
    while(rightIndex > 0){
        int low = 0, high = rightIndex;
        while(low < high){
            lists[low]= merge(lists[low], lists[high]); // 详见合并两个排序列表，https://www.cnblogs.com/MarkLeeBYR/p/16853659.html
            low++;
            high--;
        }
        rightIndex=high;
    }
    return lists[0];
}

例如有8个链表，初始rightIndex=7，

假如有8个链表，初始rightIndex=7

low=0,high=7:
　　merge(lists[0], lists[7])
　   merge(lists[1], lists[6])
　　merge(lists[2], lists[5])
　　merge(lists[3], lists[4])
　　rightIndex=3
low=0,high=3:
　　merge(lists[0], lists[3])
　　merge(lists[1], lists[2])
　　rightIndex=1
low=0,high=1:
　　merge(lists[0], lists[1])
　　rightIndex=0