Merge K Sorted List

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

11/3 这道题也跟狗家的onsite有点像，下面讲讲俺的思路：

之前的解法：

把这些lists两两merge，知道最后只剩下一个list

public class Solution {    public ListNode mergeKLists(List<ListNode> lists) {         /*CAUTION: use a loop to remove throws an error         http://stackoverflow.com/questions/8189466/java-util-concurrentmodificationexception         */         //if(lists.size() > 300)  return null;                   List<ListNode> remove = new ArrayList<ListNode>();         for(int i= lists.size()-1; i>=0; i--){             if(lists.get(i) == null){                 lists.remove(i);             }         }                             ListNode tail = null, list = null;         while( !lists.isEmpty() ){             ListNode p = lists.get(0);             ArrayList<Integer> rm = new ArrayList<Integer>();             for(int i=0; i<lists.size(); i++){                 ListNode node = lists.get(i);                 if(node.val < p.val){                     p = node;                 }             }                           for(int i=lists.size()-1; i>=0; i--){                 if(lists.get(i).val == p.val){                     rm.add(Integer.valueOf(i));                 }             }                           for(int j=0; j<rm.size(); j++){                 int index = rm.get(j).intValue();                 ListNode q = lists.get(index);                 if(list == null){                     list = q;                 }else{                     tail.next = q;                 }                 tail = q;                 q = q.next;                 if(q == null){                     lists.remove(index);                 }else{                     lists.remove(index);                     lists.add(q);                 }             }                                                   }                   return list;              }}

空间复杂度是O（1），

时间复杂度：

首先假设list 2i 和 list 2i+1 合并成list i，时间是O(m), m是最长的一个list的长度，一共有n/2个要合并，假设没有空的list

然后是list 2i和list 2i+2 合并成list 2i, 时间是O(2m), 一共是n/4个需要合并

。。。

可以知道对于某一个ListNode，会被访问lg(n)次， n是list的数量，所以时间复杂度是O(Nlgn)， N是所有node的数量

另一种解法是使用一个heap储存这些lists，每次得到min的时候都需要siftdown，所以时间复杂度是O(Nlgn)

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public class Heap{        public ListNode[] list;        public int size = 0;        public Heap(int n){            list = new ListNode[n];        }        public boolean isEmpty(){            return size == 0;        }                public void heapify(){            for(int i=size-1; i>=0; i--){                siftDown(i);            }        }                public ListNode min(){            ListNode head = list[0];            list[0] = list[0].next;            if(list[0] == null){                swap(0,size-1);                size--;                siftDown(0);            }else{                siftDown(0);            }                        head.next = null;            return head;        }                private void siftDown(int from){            int left = 2*from + 1;            int right = left + 1;                        if(left >= size)    return;                        int min = from;            if(left < size && list[left].val < list[min].val)       min = left;            if(right < size && list[right].val < list[min].val)     min = right;            swap(min,from);            if(min != from)     siftDown(min);        }                private void swap(int i, int j){            ListNode tmp = list[i];            list[i] = list[j];            list[j] = tmp;        }            }        public ListNode mergeKLists(List<ListNode> lists) {        /**         * try solve with a heap         */         Heap heap = listToHeap(lists);        if(heap.size == 0)    return null;                ListNode ans = heap.min();        ListNode prev = ans;        while(!heap.isEmpty()){            prev.next = heap.min();            prev = prev.next;        }        return ans;    }        public Heap listToHeap(List<ListNode> lists){        int len = lists.size();        Heap heap = new Heap(len);        int start = 0, end = len-1;        for(ListNode list: lists){            if(list != null){                heap.list[start++] = list;            }else{                heap.list[end--] = null;            }        }        heap.size = start;        heap.heapify();        return heap;    }        }

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