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

解法1:

  采用递归的方法,不管合并几个,归根到底还是需要两两合并。

  首先想到的是前两个先合并,然后再跟第三个合并,然后第四个。。。。但是这种做法效率不高。

  换个思路,采用分治法,对数量超过2的任务进行拆分,直到最后只有一个或两个链表再进行合并。代码如下:

/**
* 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; } else if (lists.length == 1) {
return lists[0]; } else {
ListNode res = new ListNode(0);
ListNode last = res; int mid = lists.length / 2;
ListNode one = mergeKLists(Arrays.copyOfRange(lists, 0, mid));
ListNode two = mergeKLists(Arrays.copyOfRange(lists, mid, lists.length)); while (one != null && two != null) {
if (one.val < two.val) {
last.next = one;
one = one.next;
} else {
last.next = two;
two = two.next;
}
last = last.next;
} last.next = one != null ? one : two;
return res.next;
} }
}

或者:

/**
* 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;
} int n = lists.length;
while (n > 1) {
int k = (n + 1) / 2;
for (int i = 0; i < n / 2; i++) {
lists[i] = mergeTwoLists(lists[i], lists[i + k]);
}
n = k;
}
return lists[0];
} public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode head = new ListNode(0);
ListNode last = head; while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
last.next = list1;
list1 = list1.next;
} else {
last.next = list2;
list2 = list2.next;
}
last = last.next;
} last.next = list1 != null ? list1 : list2;
return head.next;
}
}

解法2:

  采用小根堆的方法,先将k个链表的首节点加入堆中,每次会自动输出最小的节点,将该节点加入到最终结果的链表中,然后将其下一个元素(如果不为null)加入堆中,直到最终堆为空,即输出了全部元素。代码如下:

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
private Comparator<ListNode> ListNodeComparator = new Comparator<ListNode>() {
public int compare(ListNode left, ListNode right) {
if (left == null) {
return 1; // 这样操作的原因是,确保堆顶(最小节点)不会为null
} else if (right == null) {
return -1; // 这样操作的原因是,确保堆顶(最小节点)不会为null
} // 因为下面操作中确保堆中不会有null,所以这两个判断不要亦可
return left.val - right.val;
}
}; public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) {
return null;
} Queue<ListNode> heap = new PriorityQueue<ListNode>(lists.length, ListNodeComparator);
for (ListNode node : lists) {
if (node != null) {
heap.add(node);
}
} ListNode dummy = new ListNode(0);
ListNode last = dummy;
while (!heap.isEmpty()) {
last.next = heap.poll();
last = last.next;
if (last.next != null) {
heap.add(last.next);
}
}
return dummy.next;
}
}

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