[LeetCode] 023. Merge k Sorted Lists (Hard) (C++/Java/Python
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索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql) 023. Merge k Sorted Lists (Hard)链接：题目：https://oj.leetcode.com/problems/merge-k-sorted-lists/ 题意：和 021. Merge Two Sorted Lists (Easy) 类似，这次要 Merge K 个。 分析：很明显可以想到利用已完成的 Merge Two Sorted Lists 的函数来用。 这两种方法都是递归调用，都可以进行记忆化，用空间换时间，不过我不清楚会不会超空间（Memory Limit Exceed），所以就没试了~ 除用2分的思路，还有更好写的方法，就是用堆(heap)，具体就是用优先队列(Priority Queue)。 代码：C++: class Solution {
public:
ListNode *mergeKLists(vector<ListNode *> &lists) {
int sz = lists.size();
if (sz == 0)
return NULL;
while (sz > 1) {
int k = (sz + 1) / 2;
for (int i = 0; i < sz / 2; i++)
lists[i] = mergeTwoLists(lists[i],lists[i + k]);
sz = k;
}
return lists[0];
}
ListNode *mergeTwoLists(ListNode *l1,ListNode *l2) {
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
ListNode *start,*p1;
if (l1->val < l2->val) {
p1 = start = l1;
l1 = l1->next;
} else {
p1 = start = l2;
l2 = l2->next;
}
while (l1 != NULL && l2 != NULL) {
if (l1->val < l2->val) {
p1->next = l1;
p1 = l1;
l1 = l1->next;
} else {
p1->next = l2;
p1 = l2;
l2 = l2->next;
}
}
if (l1 != NULL)
p1->next = l1;
else
p1->next = l2;
return start;
}
};
Java: public class Solution {
public ListNode mergeKLists(List<ListNode> lists) {
Queue<ListNode> heap = new PriorityQueue<ListNode>(new Comparator<ListNode>(){
@Override public int compare(ListNode l1,ListNode l2) {
return l1.val - l2.val;
}
});
ListNode dummy = new ListNode(0),cur = dummy,tmp;
for (ListNode list : lists) {
if (list != null) {
heap.offer(list);
}
}
while (!heap.isEmpty()) {
tmp = heap.poll();
cur.next = tmp;
cur = cur.next;
if (tmp.next != null) {
heap.offer(tmp.next);
}
}
return dummy.next;
}
}
Python: class Solution:
# @param a list of ListNode
# @return a ListNode
def mergeKLists(self,lists):
if len(lists) == 0:
return None
if len(lists) == 1:
return lists[0]
mid = len(lists) // 2
left = self.mergeKLists(lists[:mid])
right = self.mergeKLists(lists[mid:])
# merge left and right
dummy = ListNode(0)
cur = dummy
while left or right:
if right == None or (left and left.val <= right.val):
cur.next = left
left = left.next
else:
cur.next = right
right = right.next
cur = cur.next
return dummy.next
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